CBSE · NCERT · Class 11 Maths · Chapter 7

NCERT Solutions: Class 11 Maths Chapter 7 - Binomial Theorem

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Chapter-wise NCERT intext questions and exercise answers for Binomial Theorem, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 7.1 14Miscellaneous Exercise on Chapter 7 6
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1Exercise 7.114 questions
Q.1Expand each of the expressions in Exercises 1 to 5. $(1-2x)^5$v
Solution

Using $(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ with $a=1$ and $b=-2x$, we get $1-10x+40x^2-80x^3+80x^4-32x^5$.

Answer:

$1-10x+40x^2-80x^3+80x^4-32x^5$.

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Q.2Expand each of the expressions in Exercises 1 to 5. $\left(\dfrac{2}{x}-\dfrac{x}{2}\right)^5$v
Solution

Apply the binomial theorem with $a=\dfrac2x$ and $b=-\dfrac{x}{2}$. The six terms are $\dfrac{32}{x^5}$, $-\dfrac{40}{x^3}$, $\dfrac{20}{x}$, $-5x$, $\dfrac{5x^3}{8}$ and $-\dfrac{x^5}{32}$.

Answer:

$\dfrac{32}{x^5}-\dfrac{40}{x^3}+\dfrac{20}{x}-5x+\dfrac{5x^3}{8}-\dfrac{x^5}{32}$.

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Q.3Expand each of the expressions in Exercises 1 to 5. $(2x-3)^6$v
Solution

Using the sixth-power binomial coefficients $1,6,15,20,15,6,1$ with $a=2x$ and $b=-3$ gives the stated expansion.

Answer:

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

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Q.4Expand each of the expressions in Exercises 1 to 5. $\left(\dfrac{x}{3}+\dfrac{1}{x}\right)^5$v
Solution

Use the binomial theorem with $a=\dfrac{x}{3}$ and $b=\dfrac1x$. Expanding gives $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$, which simplifies to the stated expression.

Answer:

$\dfrac{x^5}{243}+\dfrac{5x^3}{81}+\dfrac{10x}{27}+\dfrac{10}{9x}+\dfrac{5}{3x^3}+\dfrac{1}{x^5}$.

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Q.5Expand each of the expressions in Exercises 1 to 5. $\left(x+\dfrac{1}{x}\right)^6$v
Solution

Use the sixth-power coefficients $1,6,15,20,15,6,1$ with $a=x$ and $b=\dfrac1x$.

Answer:

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

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Q.6Using binomial theorem, evaluate each of the following: $(96)^3$v
Solution

$(96)^3=(100-4)^3=100^3-3\cdot100^2\cdot4+3\cdot100\cdot4^2-4^3=884736$.

Answer:

$884736$.

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Q.7Using binomial theorem, evaluate each of the following: $(102)^5$v
Solution

$(102)^5=(100+2)^5=100^5+5\cdot100^4\cdot2+10\cdot100^3\cdot2^2+10\cdot100^2\cdot2^3+5\cdot100\cdot2^4+2^5=11040808032$.

Answer:

$11040808032$.

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Q.8Using binomial theorem, evaluate each of the following: $(101)^4$v
Solution

$(101)^4=(100+1)^4=100^4+4\cdot100^3+6\cdot100^2+4\cdot100+1=104060401$.

Answer:

$104060401$.

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Q.9Using binomial theorem, evaluate each of the following: $(99)^5$v
Solution

$(99)^5=(100-1)^5=100^5-5\cdot100^4+10\cdot100^3-10\cdot100^2+5\cdot100-1=9509900499$.

Answer:

$9509900499$.

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Q.10Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or 1000.v
Solution

$(1.1)^{10000}=(1+0.1)^{10000}=1+10000(0.1)+$ other positive terms $>1001>1000$.

Answer:

$(1.1)^{10000}$ is larger.

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Q.11Find $(a + b)^4 – (a – b)^4$. Hence, evaluate $(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4$.v
Solution

Expanding, $(a+b)^4-(a-b)^4=8a^3b+8ab^3=8ab(a^2+b^2)$. Put $a=\sqrt3$ and $b=\sqrt2$: $8\sqrt6(3+2)=40\sqrt6$.

Answer:

$(a+b)^4-(a-b)^4=8ab(a^2+b^2)$. Hence $(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4=40\sqrt6$.

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Q.12Find $(x + 1)^6 + (x – 1)^6$. Hence or otherwise evaluate $(\sqrt2 + 1)^6 + (\sqrt2 – 1)^6.v
Solution

Odd-power terms cancel in the sum, giving $2x^6+30x^4+30x^2+2$. For $x=\sqrt2$, this is $2(8)+30(4)+30(2)+2=198$.

Answer:

$(x+1)^6+(x-1)^6=2x^6+30x^4+30x^2+2$. Hence the required value is $198$.

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Q.13Show that $9^{n+1} – 8n – 9$ is divisible by 64, whenever n is a positive integer.v
Solution

$9^{n+1}=9(1+8)^n=9\left(1+8n+{}^nC_2 8^2+{}^nC_3 8^3+\cdots\right)$. Therefore $9^{n+1}-8n-9=64n+9\left({}^nC_2 8^2+{}^nC_3 8^3+\cdots\right)$, which is a multiple of 64.

Answer:

$9^{n+1}-8n-9$ is divisible by $64$.

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Q.14Prove that $\sum_{r=0}^{n}3^r{}^nC_r=4^n$.v
Solution

By the binomial theorem, $(1+3)^n=\sum_{r=0}^{n}{}^nC_r1^{n-r}3^r=\sum_{r=0}^{n}3^r{}^nC_r$. Since $(1+3)^n=4^n$, the result follows.

Answer:

The identity is proved.

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2Miscellaneous Exercise on Chapter 76 questions
Q.1If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint write an = (a – b + b)n and expand]v
Solution

Write $a=(a-b)+b$. Then $a^n=((a-b)+b)^n=b^n+nC_1b^{n-1}(a-b)+nC_2b^{n-2}(a-b)^2+\cdots+(a-b)^n$. Subtracting $b^n$, every remaining term contains the factor $a-b$. Hence $a-b$ divides $a^n-b^n$.

Answer:

$a-b$ is a factor of $a^n-b^n$.

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Q.2Evaluate $(\sqrt3+\sqrt2)^6-(\sqrt3-\sqrt2)^6$.v
Solution

In $(a+b)^6-(a-b)^6$, only odd powers of $b$ remain: $12a^5b+40a^3b^3+12ab^5$. With $a=\sqrt3$ and $b=\sqrt2$, this is $108\sqrt6+240\sqrt6+48\sqrt6=396\sqrt6$.

Answer:

$396\sqrt6$.

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Q.3Find the value of $(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$.v
Solution

Let $A=a^2$ and $B=\sqrt{a^2-1}$. Then $(A+B)^4+(A-B)^4=2(A^4+6A^2B^2+B^4)$. Substituting $A=a^2$ and $B^2=a^2-1$ gives $2a^8+12a^6-10a^4-4a^2+2$.

Answer:

$2a^8+12a^6-10a^4-4a^2+2$.

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Q.4Find an approximation of $(0.99)^5$ using the first three terms of its expansion.v
Solution

$(0.99)^5=(1-0.01)^5$. Using the first three terms gives $1-5(0.01)+10(0.01)^2=1-0.05+0.001=0.951$.

Answer:

$(0.99)^5\approx0.951$.

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Q.5Expand using Binomial Theorem $\left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^4$, $x\ne0$.v
Solution

Treat the expression as $\left(1+\left(\dfrac{x}{2}-\dfrac2x\right)\right)^4$ and expand by the binomial theorem, then simplify powers of $x$. The resulting Laurent expansion is the stated expression.

Answer:

$\dfrac{x^4}{16}+\dfrac{x^3}{2}+\dfrac{x^2}{2}-4x-5+\dfrac{16}{x}+\dfrac{8}{x^2}-\dfrac{32}{x^3}+\dfrac{16}{x^4}$.

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Q.6Find the expansion of $(3x^2 – 2ax + 3a^2)^3$ using binomial theorem.v
Solution

Expand the trinomial by grouping, for example as $(3x^2+(-2ax+3a^2))^3$, and then expand the powers of $(-2ax+3a^2)$. Collecting like terms gives $27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$.

Answer:

$27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$.

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