Binomial Theorem | ISC Class 11 Maths Notes

This note covers binomials and their expansion, the history and construction of Pascal’s triangle, binomial coefficients, the binomial theorem and its proof, general and middle terms, expansions involving subtraction and fractions, coefficient identities, numerical calculations, comparisons, and divisibility.

What is a binomial, and how did its expansion develop?

What do the basic words mean?

A binomial is an algebraic expression containing two unlike terms. In a + b, the letters a and b represent the two quantities being added. A term is a part of an expression separated from other parts by addition or subtraction.

An expansion rewrites a power of an expression as a sum of terms. An index, also called an exponent, tells us the power. In (a + b)ⁿ, n is the index; here it is a positive integer, meaning a whole number greater than zero.

For small powers, direct multiplication gives familiar identities, equations valid for all permitted values of their letters. These identities show how the powers of the two quantities and their numerical multipliers change together.

A coefficient is the multiplier of a specified algebraic part. Thus, 3 is the coefficient of a²b in the cubic expansion. The number of displayed terms is one more than the index, and the two exponents in each term add to that index.

Why was a general rule useful?

Repeated multiplication becomes difficult for higher powers such as 98⁵ or 101⁶. The binomial theorem gives a systematic expansion. Its usefulness lies in predicting each coefficient and each pair of powers without carrying out all the intermediate multiplications.

The coefficient pattern has a long history. Ancient Indian mathematicians knew the coefficients for powers from zero to seven. Pingla presented their triangular arrangement as Meru-Prastara in Chhanda shastra. A triangular arrangement also appeared in the work of the Chinese mathematician Chu-shi-kie.

Michael Stipel introduced the term binomial coefficients in approximately 1544. The arithmetic triangle became popularly known as Pascal’s triangle after the French mathematician Blaise Pascal. His work on the integral form of the theorem was published posthumously, meaning after his death, in 1665.

Note: The theorem developed here has a positive integral index. Its finite expansion must not be applied unchanged to a negative or fractional index.

How does Pascal’s triangle generate binomial coefficients?

How is each row constructed?

Pascal’s triangle is an arrangement of the numerical coefficients of successive binomial expansions. Begin with 1 at the top. Put 1 at both ends of each following row. Obtain each interior entry by adding the two entries immediately above it.

The top row is labelled index zero, so the row label identifies the power being expanded. For the zero-power identity, (a + b)⁰ = 1, the base a + b must be non-zero. The later rows give the coefficients for positive powers.

The diagram has rows labelled with indices 0 to 4, with 1 along the two sloping sides. Blue triangular markers between selected entries illustrate the addition that produces interior entries in the following row.

See Fig. 7.2 in your NCERT textbook

These entries are binomial coefficients. They provide the numerical pattern before the actual quantities are substituted. In an expansion involving 2x and 3y, where x and y are variables representing numbers, the powers of 2 and 3 also contribute to the final coefficients.

How do combinations give a row directly?

The notation $\binom{n}{r}$, read “n choose r”, counts selections of r objects from n distinct objects when order does not matter. Here n is a non-negative integer and r is an integer from zero to n. Such a selection is a combination.

The symbol n! means factorial: the product of the positive integers from 1 to n, with 0! defined as 1. The coefficient formula is $\binom{n}{r}=\frac{n!}{r!(n-r)!}$. In particular, $\binom{n}{0}=\binom{n}{n}=1$.

Rows labelled 0 to 5 display combination symbols with their numerical values underneath. The bottom row has the values 1, 5, 10, 10, 5, 1.

See Fig. 7.3 in your NCERT textbook

Property: Pascal’s addition rule

For a positive integer k and an integer r from 1 to k, $\binom{k}{r}+\binom{k}{r-1}=\binom{k+1}{r}$. The letter k labels one row; k + 1 labels the next. This identity expresses the same addition used to build the triangle.

Constructing every row up to index 12 is a slightly lengthy process. Combination notation supplies a chosen row directly, without requiring all earlier rows. It also explains how the coefficient pattern continues for an arbitrary positive integral index.

What does the binomial theorem state, and how is it proved?

Theorem: Expansion for a positive integral index

For quantities a and b and a positive integer n, the binomial theorem states that $(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$. The symbol ∑ means “sum”; here it instructs us to add the expressions for every integer r from zero to n.

Written out, the expansion begins with aⁿ, followed by naⁿ⁻¹b, and ends with nabⁿ⁻¹ followed by bⁿ. The coefficient of aⁿ⁻ʳbʳ is $\binom{n}{r}$. Each term uses one coefficient, one power of a and one power of b.

There are n + 1 terms in the standard expansion. The exponent of a falls from n to zero, while that of b rises from zero to n. Their sum is n in every term. These observations provide useful checks on a completed expansion.

How does mathematical induction establish the formula?

Mathematical induction proves a statement for every positive integer by establishing its first case and showing that each case implies the next. Let P(n) name the proposed binomial expansion with index n. Let k be an arbitrary positive integer.

1. Base case: For n = 1, the right side is $\binom{1}{0}a+\binom{1}{1}b=a+b$. It equals (a + b)¹, so P(1) is true.
2. Induction hypothesis: Assume P(k) is true. This means $(a+b)^k=\sum_{r=0}^{k}\binom{k}{r}a^{k-r}b^r$. The assumption is made for one arbitrary positive integer k.
3. Multiply: Multiply the assumed expansion by a + b. Multiplication by a produces one group of terms, while multiplication by b produces another group.
4. Combine like terms: Like terms have the same algebraic factors and powers. For an interior term aᵏ⁺¹⁻ʳbʳ, where 1 ≤ r ≤ k and ≤ means “less than or equal to”, the combined coefficient is $\binom{k}{r}+\binom{k}{r-1}$.
5. Use Pascal’s rule: Replace this sum by $\binom{k+1}{r}$. The two end terms are aᵏ⁺¹ and bᵏ⁺¹, each with coefficient 1.
6. Conclude: The resulting expression is exactly P(k + 1). Since P(1) is true and P(k) implies P(k + 1), the theorem holds for every positive integer n.

The key step is the addition of neighbouring coefficients. The proof therefore connects multiplication of binomials with the construction of Pascal’s triangle. The endpoint coefficients need separate attention because each endpoint comes from just one of the two multiplied groups.

The induction hypothesis is not the conclusion being assumed for every n. It is the temporary assumption used to demonstrate the transition from index k to index k + 1. Together with the verified first case, that transition completes the proof.

How are complete expansions written without losing factors or signs?

What should be substituted first?

Identify the two complete quantities inside the brackets before expanding. Treat each as a single unit until its required power has been written. Then simplify numerical powers and variable powers separately. This prevents coefficients inside the original binomial from being lost.

Worked example 1. Expand (2x + 3y)⁵, where x and y are variables.

Use the coefficient row 1, 5, 10, 10, 5, 1. Before simplification the expression is (2x)⁵ + 5(2x)⁴(3y) + 10(2x)³(3y)² + 10(2x)²(3y)³ + 5(2x)(3y)⁴ + (3y)⁵.

Answer: 32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵.

There are six terms, as required by index five. In every term the exponents of x and y add to five. The first and last coefficients are 32 and 243 because the original quantities include 2 and 3.

Worked example 2. Expand (x + 2)⁶.

The coefficients of the binomial pattern are 1, 6, 15, 20, 15, 6, 1. Multiply them respectively by 1, 2, 2², 2³, 2⁴, 2⁵ and 2⁶ while the power of x decreases.

Answer: x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64.

How is subtraction handled?

For a difference, the second quantity includes its minus sign. Thus (x − y)ⁿ means [x + (−y)]ⁿ. For odd r, (−y)ʳ = −yʳ; for even r, (−y)ʳ = yʳ. These factors produce alternating signs in the displayed expansion.

Worked example 3. Expand (x − 2y)⁵.

Use x as the first quantity and −2y as the second. The unsimplified terms are x⁵ + 5x⁴(−2y) + 10x³(−2y)² + 10x²(−2y)³ + 5x(−2y)⁴ + (−2y)⁵.

Answer: x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵.

What changes when a quantity contains a fraction?

The theorem still applies when a quantity contains a fraction, provided the denominator, the quantity below the fraction bar, is non-zero. This domain restriction specifies which variable values are permitted. Powers now act on both the numerator, the quantity above the fraction bar, and the denominator before like powers of the variable are simplified.

Worked example 4. Expand (x² + 3/x)⁴ for x ≠ 0, where ≠ means “not equal to”.

The five terms are (x²)⁴ + 4(x²)³(3/x) + 6(x²)²(3/x)² + 4x²(3/x)³ + (3/x)⁴.

Answer: x⁸ + 12x⁵ + 54x² + 108/x + 81/x⁴, with x ≠ 0.

The two original quantities still have exponents adding to four before simplification. The simplified powers of x do not follow the ordinary descending-by-one pattern, because both original quantities themselves contain x. Distinguishing these stages prevents a misleading check.

How do the general term and a required coefficient differ?

Result: The general term

The general term is a formula that identifies any term in the expansion by its position. Write T with a subscript to denote a term; for example, T₁ means the first term. Then $T_{r+1}=\binom{n}{r}a^{n-r}b^r$, for integers r from zero to n.

The position is r + 1, because the first term corresponds to r = 0. To find a specified term, subtract one from its position to obtain r. Substituting the position itself for r gives the following term and is a common indexing error.

A required coefficient is the multiplier of a requested power or algebraic part. Finding it involves matching the power after substitution and simplification. The term position and the exponent of a variable need not be the same number.

Worked example 5. Find the fourth term and the coefficient of x³ in (x + 2)⁶.

For the fourth term, r + 1 = 4, so r = 3. Substitute a = x, b = 2 and n = 6 into the general term.

Answer: $T_4=\binom{6}{3}x^3 2^3=160x^3$. The fourth term is 160x³, and the coefficient of x³ is 160.

The binomial coefficient in this calculation is 20, whereas the coefficient of x³ is 160. The factor 2³ accounts for the difference. Calling both numbers simply “the coefficient” without specifying the algebraic part would hide an essential distinction.

How should powers be matched after substitution?

For (x² + 3/x)⁴, with x ≠ 0, the general term simplifies to $T_{r+1}=\binom{4}{r}3^r x^{8-3r}$. The exponent 8 − 3r comes from 2(4 − r) − r: the numerator contributes one power and the denominator subtracts another.

To locate the term containing x², solve 8 − 3r = 2. This gives r = 2, so the required term is third. Its coefficient is $\binom{4}{2}3^2=54$, agreeing with the complete expansion.

A term independent of x has exponent zero after simplification. Therefore, set the simplified exponent equal to zero and check whether the resulting r is an allowed integer. A fractional r does not identify a term in a finite binomial expansion.

In the same fourth-power expansion, 8 − 3r = 0 gives r = 8/3. Since this is not an integer, there is no term independent of x. The general term can establish this without requiring a separate search through the expansion.

How are the middle term or middle terms located?

Result: Middle position depends on the number of terms

A middle term occupies the central position in the ordered binomial expansion. Begin by counting n + 1 terms. The parity of n, meaning whether it is even or odd, determines whether that count has one central position or two.

If n is even, n + 1 is odd, so there is one middle term, at position n/2 + 1. Consequently, $T_{n/2+1}=\binom{n}{n/2}a^{n/2}b^{n/2}$. This follows directly by putting r = n/2 in the general term.

If n is odd, n + 1 is even, so there are two middle terms. Their positions are (n + 1)/2 and (n + 3)/2. Their corresponding r values are (n − 1)/2 and (n + 1)/2, respectively.

How do positions become actual terms?

Worked example 6. Find the middle term of (x + 2)⁶.

There are seven terms, so the fourth is central. Set r = 3 in the general term. This gives $T_4=\binom{6}{3}x^3 2^3$.

Answer: The middle term is 160x³. The factor 2³ must be included even though the binomial coefficient itself is 20.

Worked example 7. Find the middle terms of (2x + 3y)⁵.

There are six terms, so the third and fourth are central. Their r values are 2 and 3. Use $T_3=\binom{5}{2}(2x)^3(3y)^2$ and $T_4=\binom{5}{3}(2x)^2(3y)^3$.

Answer: The middle terms are 720x³y² and 1080x²y³, in that order.

The middle position is determined by the index, but the actual term depends on both quantities inside the binomial. Equal central binomial coefficients do not make the corresponding complete terms equal. Here both central binomial coefficients are 10, yet their other factors differ.

Keep the original order when naming terms. Start with the highest power of the first quantity and progress towards the highest power of the second. This makes the general-term index and the middle-position rule consistent with the same expansion.

These rules concern position, rather than the numerical size of a term after values are assigned to variables. Finding a central position requires the number of terms; it does not require comparing their values.

What identities follow from special binomial expansions?

How do the expansions of 1 + x and 1 − x help?

When the first quantity is 1, its powers disappear from the written expression. Thus $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^r$. Here x is a variable, n is a positive integer and r runs through the integers from zero to n.

The coefficients in this particular expansion are exactly the binomial coefficients. Substituting a convenient value of x turns the expansion into a numerical identity involving a whole row of Pascal’s triangle. This avoids calculating each coefficient separately.

Identity: Sum of binomial coefficients

Set x = 1 in the expansion of (1 + x)ⁿ. Every power of x is then 1, so $\sum_{r=0}^{n}\binom{n}{r}=2^n$. This is the sum of all the binomial coefficients for index n, including both endpoint coefficients.

The identity concerns the standard binomial coefficient row. In an expansion such as (x + 2)⁶, powers of 2 also multiply that row. Its final coefficients therefore form a different list, so the row-sum identity must be applied with care.

Identity: Alternating sum of binomial coefficients

Replacing x by −x gives the expansion of (1 − x)ⁿ. Now put x = 1. Because n is positive, the left side is zero. Hence $\sum_{r=0}^{n}(-1)^r\binom{n}{r}=0$.

An alternating sum adds successive entries with alternating plus and minus signs. In this identity, r starts at zero, so the first coefficient is positive. The zero result expresses a balance between the coefficients carrying positive signs and those carrying negative signs.

Worked example 8. Prove $\sum_{r=0}^{n}\binom{n}{r}3^r=4^n$ for a positive integer n.

Use (1 + x)ⁿ and substitute x = 3. The general expression on the right becomes $\binom{n}{r}3^r$, which is exactly the required summand, meaning the expression being added.

Answer: $\sum_{r=0}^{n}\binom{n}{r}3^r=(1+3)^n=4^n$.

In this example, each coefficient is multiplied by a power of 3 before addition. Recognising that factor suggests the substitution immediately. Read the index range as well as the coefficient: omitting an endpoint would no longer reproduce the full binomial expansion.

The same method links a long sum to a short power. First identify the binomial coefficient, then identify the accompanying power, and finally choose the two quantities whose expansion produces both. The justification is substitution into the theorem, not a numerical guess.

How does the theorem support numerical calculations, comparisons and divisibility?

How can a nearby convenient number simplify a power?

For numerical evaluation, rewrite a number as a sum or difference whose powers are easier to calculate. Use the complete expansion when an exact answer is required. Keeping the signs attached to the terms is especially useful when the second quantity is negative.

Worked example 9. Calculate 98⁵ using the binomial theorem.

Write 98 = 100 − 2. Then 98⁵ = 100⁵ − 5(100)⁴(2) + 10(100)³(2²) − 10(100)²(2³) + 5(100)(2⁴) − 2⁵.

Group the positive and negative contributions: 98⁵ = 10040008000 − 1000800032.

Answer: 98⁵ = 9039207968.

When is a partial expansion sufficient?

For a comparison, an exact value may be unnecessary. If the omitted terms are positive, the displayed terms give a lower bound, meaning a value below the full sum. The signs of the omitted terms are what justify the inequality.

Worked example 10. Which is larger, (1.01)¹⁰⁰⁰⁰⁰⁰ or 10000?

Write (1.01)¹⁰⁰⁰⁰⁰⁰ = (1 + 0.01)¹⁰⁰⁰⁰⁰⁰. Its expansion begins with 1 + 1000000 × 0.01, followed by other positive terms.

Answer: (1.01)¹⁰⁰⁰⁰⁰⁰ = 1 + 10000 + other positive terms > 10000. Therefore the power is larger.

A partial expansion here proves the comparison exactly; it is not being presented as an exact evaluation of the power. In a difference expansion, terms have alternating signs, so simply discarding the remaining terms would need a separate justification.

How does expansion reveal a remainder?

A number is divisible by a positive integer when it is an integer multiple of that integer. In division, the quotient is the integer multiplier and the remainder is what remains, from zero up to one less than the divisor.

Worked example 11. Show that 6ⁿ − 5n leaves remainder 1 on division by 25 for every positive integer n. Here 5n means five multiplied by n.

For n = 1, 6 − 5 = 1. For n ≥ 2, expand 6ⁿ = (1 + 5)ⁿ. Its first two terms are 1 + 5n; every later term contains a factor 5² = 25.

Answer: 6ⁿ − 5n = 1 + 25q, where $q=\sum_{r=2}^{n}\binom{n}{r}5^{r-2}$ is an integer. Thus the remainder is 1. For n = 1, q = 0.

The linear term, the term containing the first power of 5, cancels with 5n. All remaining non-constant terms contain the required divisor. Isolating those first two terms is enough; calculating the later coefficients individually would add unnecessary work.

Glossary

• Binomial — An algebraic expression containing two unlike terms joined by addition or subtraction.
• Expansion — A sum of terms obtained by multiplying out a power of an expression.
• Index — The exponent specifying the power to which a quantity or expression is raised.
• Coefficient — The multiplier of a specified variable power or algebraic part of a term.
• Factorial — The product of positive integers up to a given positive integer, with zero factorial defined as one.
• Combination — A selection of objects in which the order of selection does not matter.
• Binomial coefficient — A combination number giving the coefficient of a specified product of powers in the standard binomial expansion.
• Pascal’s triangle — A triangular coefficient array with endpoint entries one and interior entries formed by adding neighbouring entries above.
• Mathematical induction — A proof method establishing a first case and showing that each case implies the next.
• General term — A formula specifying any term in an expansion through an index linked to its position.
• Middle term — A term occupying a central position in the ordered terms of a binomial expansion.
• Independent term — A term whose simplified exponent of the specified variable is zero.
• Remainder — The amount left after subtracting the divisor multiplied by the integer quotient.

Common errors and misconceptions

• Misconception: Index n gives n terms. Correct: The standard expansion has n + 1 terms, including the first and last terms.
• Misconception: The fourth term uses r = 4. Correct: The general term is numbered r + 1, so the fourth term uses r = 3.
• Misconception: An even index gives two middle terms. Correct: An even index gives an odd number of terms and one middle term; an odd index gives two.
• Misconception: The binomial coefficient is the complete coefficient after substitution. Correct: Powers of numerical factors also contribute, as in the coefficient 160 of x³ in (x + 2)⁶.
• Misconception: Every term after the first is negative in a difference expansion. Correct: Powers of the negative second quantity produce alternating signs.
• Misconception: Any solution for r locates an independent term. Correct: The solution must be an integer from zero to n; a fractional solution gives no term.
• Misconception: A fractional expression can be expanded without restrictions. Correct: A denominator must remain non-zero; (x² + 3/x)⁴ requires x ≠ 0.
• Misconception: A few displayed terms equal the full expansion. Correct: In the comparison involving (1.01)¹⁰⁰⁰⁰⁰⁰, the omitted terms are positive, establishing an inequality rather than equality with the partial sum.

Exam-style questions with model answers

Q1. State the number of terms and describe the powers of a and b in the standard expansion of (a + b)ⁿ, where n is a positive integer and a and b are the two quantities. [2 marks]

1. The expansion contains n + 1 terms, one more than its positive integral index.
2. The exponent of a decreases from n to zero while that of b increases from zero to n; their sum is n in each term.

Q2. Find the fourth term of (x + 2)⁶ and state the coefficient of x³, where x is a variable. [3 marks]

1. Write the general term as $T_{r+1}=\binom{6}{r}x^{6-r}2^r$, where T denotes a term and r is an integer from zero to six. The fourth position requires r = 3.
2. Substitution gives $T_4=\binom{6}{3}x^3 2^3$. Evaluate the binomial coefficient as 20 and the numerical power as 8.
3. Multiplying gives T₄ = 160x³. Therefore the required term is 160x³ and the coefficient of x³ is 160.

Q3. Find the middle terms of (2x + 3y)⁵, where x and y are variables. [4 marks]

1. The expansion has six terms. Its two central positions are the third and fourth, since the index five is odd.
2. In the general term $T_{r+1}=\binom{5}{r}(2x)^{5-r}(3y)^r$, T denotes a term and r is the integer one less than its position. Thus use r = 2 and r = 3.
3. The third term is $\binom{5}{2}(2x)^3(3y)^2=10\times8x^3\times9y^2=720x^3y^2$.
4. The fourth term is $\binom{5}{3}(2x)^2(3y)^3=10\times4x^2\times27y^3=1080x^2y^3$. These are the two middle terms.

Q4. Prove the binomial theorem $(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$ by mathematical induction for positive integers n. Here a and b are quantities, r is the summation index, ∑ means summation, and the bracketed coefficient means “n choose r”. You may use Pascal’s identity $\binom{k}{r}+\binom{k}{r-1}=\binom{k+1}{r}$ for integers 1 ≤ r ≤ k. [6 marks]

1. Let P(n) denote the stated expansion. For n = 1, the right side is a + b because both binomial coefficients equal one. Thus the base case P(1) holds.
2. Assume P(k) holds for an arbitrary positive integer k. Hence the expansion of (a + b)ᵏ has coefficient $\binom{k}{r}$ on aᵏ⁻ʳbʳ.
3. Multiply the assumed expansion by a + b. Distribute a over every term and then distribute b over every term.
4. For each interior power product aᵏ⁺¹⁻ʳbʳ, the contributions combine to give coefficient $\binom{k}{r}+\binom{k}{r-1}$, where 1 ≤ r ≤ k.
5. Pascal’s identity changes this coefficient to $\binom{k+1}{r}$. The first and last terms are aᵏ⁺¹ and bᵏ⁺¹, with coefficient one.
6. The resulting expression is precisely P(k + 1). The true base case and established induction step prove the theorem for every positive integer n.

Q5. Expand (x² + 3/x)⁴, where x is a non-zero variable, and determine whether its expansion contains a term independent of x. [5 marks]

1. Treat x² and 3/x as the two complete quantities. For the fourth power, the binomial coefficient row is 1, 4, 6, 4, 1.
2. The first three terms simplify to (x²)⁴ = x⁸, 4(x²)³(3/x) = 12x⁵ and 6(x²)²(3/x)² = 54x².
3. The remaining terms are 4x²(3/x)³ = 108/x and (3/x)⁴ = 81/x⁴. Therefore the full expansion is x⁸ + 12x⁵ + 54x² + 108/x + 81/x⁴.
4. For an integer index r from zero to four, the general term contains the power x⁸⁻³ʳ. Independence from x requires its exponent to be zero.
5. Solving 8 − 3r = 0 gives r = 8/3, which is not an allowed integer. Thus no term is independent of x; the restriction x ≠ 0 remains.

Q6. Use the binomial theorem to decide which is larger: (1.01)¹⁰⁰⁰⁰⁰⁰ or 10000. Explain why the uncalculated terms do not invalidate the comparison. [3 marks]

1. Rewrite the power as (1 + 0.01)¹⁰⁰⁰⁰⁰⁰. The first two terms of its binomial expansion are 1 and 1000000 × 0.01.
2. The second term equals 10000. Every remaining term is positive because its binomial coefficient is positive and its factors are powers of positive numbers.
3. The full value is therefore 1 + 10000 plus other positive terms, which is greater than 10000. Hence (1.01)¹⁰⁰⁰⁰⁰⁰ is the larger number.

Q7. For every positive integer n, prove that 6ⁿ − 5n leaves remainder 1 when divided by 25. The expression 5n means five multiplied by n. [5 marks]

1. For n = 1, the expression equals 6 − 5 = 1, which has the required remainder. Consider n ≥ 2 for the remaining argument.
2. Write 6ⁿ = (1 + 5)ⁿ. Its binomial expansion begins with 1 + 5n; subsequent terms have the form $\binom{n}{r}5^r$, for integers r from 2 to n.
3. Subtract 5n from the expansion. The linear contribution cancels, leaving the constant 1 together with all terms whose powers of 5 are at least two.
4. Each of those later terms is divisible by 5² = 25, since its binomial coefficient is an integer. Their sum can therefore be written as 25q for an integer q.
5. Thus 6ⁿ − 5n = 25q + 1. As 1 is non-negative and less than 25, the remainder is 1, completing the proof.

Key takeaways

• The binomial theorem gives a finite expansion for a positive integral index, with one more term than the index.
• Pascal’s triangle builds each interior coefficient from two neighbouring entries above, while both endpoint entries remain one.
• The combination formula gives binomial coefficients directly, removing the need to construct every earlier row of the triangle.
• The general term uses index r for position r + 1; distinguish the complete term from its numerical coefficient.
• An even index gives one middle term, whereas an odd index gives two; count the terms before substituting.
• For a difference expansion, keep the negative sign inside the second quantity before raising it to successive powers.
• Substituting one into the special expansions gives the sum and alternating sum identities for binomial coefficients.
• Positive omitted terms can establish a comparison, while common factors in later terms can establish a divisibility result.

Test yourself

How many terms occur in the standard expansion of (a + b)ⁿ for positive integer n?

There are n + 1 terms, corresponding to integer index values from zero to n.

How is an interior entry of Pascal’s triangle obtained?

Add the two neighbouring entries immediately above it; the endpoint entries of each row are one.

Which value of r identifies the fourth term when the general term is written as T with subscript r + 1?

Use r = 3, because the term’s position is one greater than r.

Why does (x + 2)⁶ have one middle term?

It contains seven terms, so its fourth term occupies the single central position.

What are the middle terms of (2x + 3y)⁵?

The third and fourth terms are central: 720x³y² and 1080x²y³, respectively.

What is the sum of the binomial coefficients for a positive integral index n?

The sum is 2ⁿ, obtained by substituting one for the variable in (1 + x)ⁿ.

Does (x² + 3/x)⁴, with x ≠ 0, contain an independent term?

No. Its exponent condition gives r = 8/3, which is not an integer index of a term.

Why do later terms in the expansion of (1 + 5)ⁿ help establish a remainder on division by 25?

After the constant and linear terms, every term contains at least 5², so each is divisible by 25.