NCERT Class 11 Mathematics — Binomial Theorem, Exercise 7.1. All 14 questions solved.

The theorem itself, for any positive integer $n$:

$$(a + b)^n = {}^n\mathrm{C}_0 a^n + {}^n\mathrm{C}_1 a^{n-1}b + {}^n\mathrm{C}_2 a^{n-2}b^2 + \cdots + {}^n\mathrm{C}_n b^n$$

Three features are worth fixing in mind before starting:

• There are $n + 1$ terms, and the powers of $a$ fall from $n$ to $0$ while those of $b$ rise from $0$ to $n$ — their sum is always $n$, which is the quickest check on any term.
• The coefficients $^n\mathrm{C}_r$ are symmetric: $^n\mathrm{C}_r = {}^n\mathrm{C}_{n-r}$.
• If $b$ is negative, the signs alternate, because $(-b)^r$ is negative for odd $r$.

For $n = 5$ and $n = 6$ the coefficient rows from Pascal’s triangle are

$$1,\ 5,\ 10,\ 10,\ 5,\ 1 \qquad\text{and}\qquad 1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$$

Key insight. Questions 6 to 9 are the theorem used as an arithmetic device: $(102)^5$ is hard, but $(100 + 2)^5$ is a sum of six terms each of which is a power of $100$ times a small number. Choosing the split so that one part is a round number — $96 = 100 – 4$, $99 = 100 – 1$ — is what makes the method pay.

Expand each of the expressions in questions 1 to 5.

Question 1

$(1 – 2x)^5$

Solution. Take $a = 1$, $b = -2x$, $n = 5$. Since $b$ is negative the signs alternate, and each term carries $(-2)^r$:

$$(1-2x)^5 = \sum_{r=0}^{5} {}^5\mathrm{C}_r (1)^{5-r}(-2x)^r$$

Term by term, with coefficients $1, 5, 10, 10, 5, 1$:

$$= 1 – 5(2x) + 10(4x^2) – 10(8x^3) + 5(16x^4) – 32x^5$$

$$= 1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5$$

Question 2

$\left(\dfrac{2}{x} – \dfrac{x}{2}\right)^5$

Solution. Here $a = \dfrac2x$ and $b = -\dfrac{x}{2}$. Note that $a$ carries a negative power of $x$, so the powers of $x$ in the answer run from $-5$ up to $+5$ in steps of $2$.

$$= {}^5\mathrm{C}_0\left(\frac2x\right)^5 – {}^5\mathrm{C}_1\left(\frac2x\right)^4\left(\frac x2\right) + {}^5\mathrm{C}_2\left(\frac2x\right)^3\left(\frac x2\right)^2 – \cdots$$

Evaluating each term:

$$\left(\frac2x\right)^5 = \frac{32}{x^5}, \qquad 5\left(\frac{16}{x^4}\right)\left(\frac x2\right) = \frac{40}{x^3}$$

$$10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right) = \frac{20}{x}, \qquad 10\left(\frac{4}{x^2}\right)\left(\frac{x^3}{8}\right) = 5x$$

$$5\left(\frac2x\right)\left(\frac{x^4}{16}\right) = \frac{5x^3}{8}, \qquad \left(\frac x2\right)^5 = \frac{x^5}{32}$$

With alternating signs:

$$\frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5x^3}{8} – \frac{x^5}{32}$$

Question 3

$(2x – 3)^6$

Solution. $a = 2x$, $b = -3$, $n = 6$, coefficients $1, 6, 15, 20, 15, 6, 1$:

$$(2x)^6 = 64x^6$$ $$6(2x)^5(3) = 6 \times 32x^5 \times 3 = 576x^5$$ $$15(2x)^4(9) = 15 \times 16x^4 \times 9 = 2160x^4$$ $$20(2x)^3(27) = 20 \times 8x^3 \times 27 = 4320x^3$$ $$15(2x)^2(81) = 15 \times 4x^2 \times 81 = 4860x^2$$ $$6(2x)(243) = 2916x$$ $$3^6 = 729$$

Attaching the alternating signs:

$$64x^6 – 576x^5 + 2160x^4 – 4320x^3 + 4860x^2 – 2916x + 729$$

Question 4

$\left(\dfrac{x}{3} + \dfrac{1}{x}\right)^5$

Solution. All signs are positive here. With $a = \dfrac x3$, $b = \dfrac1x$:

$$\left(\frac x3\right)^5 = \frac{x^5}{243}$$ $$5\left(\frac{x^4}{81}\right)\left(\frac1x\right) = \frac{5x^3}{81}$$ $$10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right) = \frac{10x}{27}$$ $$10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right) = \frac{10}{9x}$$ $$5\left(\frac x3\right)\left(\frac{1}{x^4}\right) = \frac{5}{3x^3}$$ $$\left(\frac1x\right)^5 = \frac{1}{x^5}$$

Question 5

$\left(x + \dfrac{1}{x}\right)^6$

Solution. Each term is $^6\mathrm{C}_r\,x^{6-r}\,x^{-r} = {}^6\mathrm{C}_r\,x^{6-2r}$, so the powers step down by $2$ from $6$ to $-6$:

$$x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$

The expansion is palindromic — it reads the same forwards and backwards — because replacing $x$ by $\tfrac1x$ leaves the original expression unchanged.

Using the binomial theorem, evaluate each of the following.

Question 6

$(96)^3$

Solution. Write $96 = 100 – 4$ and expand with $n = 3$, coefficients $1, 3, 3, 1$:

$$(100 – 4)^3 = 100^3 – 3(100)^2(4) + 3(100)(4)^2 – 4^3$$

$$= 1000000 – 120000 + 4800 – 64 = 884736$$

Question 7

$(102)^5$

Solution. $102 = 100 + 2$, $n = 5$, coefficients $1, 5, 10, 10, 5, 1$:

$$(100+2)^5 = 100^5 + 5(100)^4(2) + 10(100)^3(4) + 10(100)^2(8) + 5(100)(16) + 32$$

$$= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$$

$$= 11040808032$$

Question 8

$(101)^4$

Solution. $101 = 100 + 1$, so every power of $1$ is $1$ and the terms are just the coefficients times powers of $100$:

$$100^4 + 4(100)^3 + 6(100)^2 + 4(100) + 1 = 100000000 + 4000000 + 60000 + 400 + 1$$

$$= 104060401$$

The digits of the answer are the binomial coefficients $1, 4, 6, 4, 1$ separated into pairs — a pleasant check that comes free with base $100$.

Question 9

$(99)^5$

Solution. $99 = 100 – 1$, so the signs alternate:

$$100^5 – 5(100)^4 + 10(100)^3 – 10(100)^2 + 5(100) – 1$$

$$= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1$$

$$= 9509900499$$

Question 10

Using the binomial theorem, indicate which number is larger, $(1.1)^{10000}$ or $1000$.

Solution. Write $1.1 = 1 + 0.1$ and expand. Every term of $(1 + 0.1)^{10000}$ is positive, so keeping just the first two gives a valid lower bound:

$$(1.1)^{10000} = {}^{10000}\mathrm{C}_0 + {}^{10000}\mathrm{C}_1(0.1) + \text{other positive terms}$$

$$= 1 + 10000 \times 0.1 + \text{other positive terms}$$

$$= 1 + 1000 + \text{other positive terms} > 1000$$

Hence

$$(1.1)^{10000} > 1000$$

Note what makes this rigorous: the discarded terms are all positive, so dropping them can only decrease the value. Two terms suffice — there is no need to estimate the rest.

Question 11

Find $(a+b)^4 – (a-b)^4$. Hence evaluate $(\sqrt3 + \sqrt2)^4 – (\sqrt3 – \sqrt2)^4$.

Solution. Expand both, noting that $(a-b)^4$ differs only in the signs of the odd powers of $b$:

$$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$ $$(a-b)^4 = a^4 – 4a^3b + 6a^2b^2 – 4ab^3 + b^4$$

Subtracting, the even-power terms cancel and the odd-power terms double:

$$(a+b)^4 – (a-b)^4 = 8a^3b + 8ab^3 = 8ab(a^2 + b^2)$$

Now substitute $a = \sqrt3$, $b = \sqrt2$, so $ab = \sqrt6$ and $a^2 + b^2 = 5$:

$$8\sqrt6 \times 5 = 40\sqrt6$$

Question 12

Find $(x+1)^6 + (x-1)^6$. Hence, or otherwise, evaluate $(\sqrt2 + 1)^6 + (\sqrt2 – 1)^6$.

Solution. This time adding cancels the odd-power terms and doubles the even ones:

$$(x+1)^6 + (x-1)^6 = 2\left(x^6 + 15x^4 + 15x^2 + 1\right)$$

Substituting $x = \sqrt2$, so $x^2 = 2$, $x^4 = 4$, $x^6 = 8$:

$$2(8 + 15 \times 4 + 15 \times 2 + 1) = 2(8 + 60 + 30 + 1) = 2 \times 99 = 198$$

The whole point of pairing the two surds is that the irrational parts cancel and an integer remains.

Question 13

Show that $9^{n+1} – 8n – 9$ is divisible by $64$, whenever $n$ is a positive integer.

Solution. Write $9 = 1 + 8$, so that powers of $8$ appear — and $64 = 8^2$ is what we are aiming to factor out.

$$9^{n+1} = (1 + 8)^{n+1} = {}^{n+1}\mathrm{C}_0 + {}^{n+1}\mathrm{C}_1(8) + {}^{n+1}\mathrm{C}_2(8)^2 + \cdots + {}^{n+1}\mathrm{C}_{n+1}(8)^{n+1}$$

The first two terms are $1$ and $8(n+1) = 8n + 8$, so

$$9^{n+1} = 1 + 8n + 8 + {}^{n+1}\mathrm{C}_2 8^2 + {}^{n+1}\mathrm{C}_3 8^3 + \cdots$$

$$9^{n+1} = 9 + 8n + 8^2\left[{}^{n+1}\mathrm{C}_2 + {}^{n+1}\mathrm{C}_3(8) + \cdots\right]$$

Therefore

$$9^{n+1} – 8n – 9 = 64\left[{}^{n+1}\mathrm{C}_2 + {}^{n+1}\mathrm{C}_3(8) + \cdots + 8^{n-1}\right]$$

The bracket is an integer, so the whole expression is $64$ times an integer, that is, divisible by $64$. $\blacksquare$

The design of the question is now visible: the subtracted $8n + 9$ is exactly the first two terms of the expansion, leaving everything from $8^2$ onwards.

Question 14

Prove that $\displaystyle\sum_{r=0}^{n} 3^r\,{}^n\mathrm{C}_r = 4^n$.

Solution. Compare the sum with the binomial expansion of $(a+b)^n$. Writing it out:

$$\sum_{r=0}^{n} {}^n\mathrm{C}_r\,3^r = {}^n\mathrm{C}_0 + {}^n\mathrm{C}_1 3 + {}^n\mathrm{C}_2 3^2 + \cdots + {}^n\mathrm{C}_n 3^n$$

This is exactly $(1 + 3)^n$, since taking $a = 1$ and $b = 3$ gives $^n\mathrm{C}_r\,1^{n-r}3^r = {}^n\mathrm{C}_r 3^r$ for the general term. Hence

$$\sum_{r=0}^{n} 3^r\,{}^n\mathrm{C}_r = (1+3)^n = 4^n \qquad \blacksquare$$

Common mistakes

• Losing the alternating signs. In questions 1, 2, 3, 6 and 9 the second term of the binomial is negative, so odd-numbered terms change sign. Writing $b = -2x$ explicitly, rather than “$2x$ with signs added later”, prevents this.
• Forgetting to raise the whole term to the power. In question 3, the fourth term needs $(2x)^3 = 8x^3$, not $2x^3$. The coefficient inside the bracket is raised along with $x$.
• Question 2, mishandling negative powers. $\left(\tfrac2x\right)^5$ is $\tfrac{32}{x^5}$, and the powers of $x$ in the answer step upwards by $2$, from $x^{-5}$ to $x^{5}$. Checking that the exponents form an arithmetic progression catches most slips.
• Questions 6 to 9, choosing an awkward split. $96 = 90 + 6$ works but leaves three-digit powers to compute. Splitting against the nearest power of $10$ is what makes the arithmetic manageable.
• Question 10, trying to evaluate $(1.1)^{10000}$. The point is a bound, not a value. Two terms already exceed $1000$, and every discarded term is positive.
• Questions 11 and 12, expanding both brackets in full and subtracting term by term. Recognising that subtraction kills the even powers, and addition kills the odd ones, halves the work and eliminates sign errors.
• Question 13, expanding $9^{n+1}$ as $9 \cdot 9^n$. That leads nowhere; the move is $9 = 1 + 8$, so that $64 = 8^2$ appears naturally from the third term onwards.
• Question 14, trying to sum the series directly. It is a binomial expansion in disguise. Matching it against $(a+b)^n$ and reading off $a = 1$, $b = 3$ is the entire proof.

Practise next

• Miscellaneous Exercise on Chapter 7 — trinomial expansions, approximations, and the same conjugate-pairing trick used in questions 11 and 12.
• Exercise 6.4 — worth revising if the coefficients $^n\mathrm{C}_r$ still need computing from scratch each time.