Exercise 7.1 (Revised) - Chapter 8 - Binomial Theorem - Ncert Solutions class 11 - Maths

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Chapter 7 - Binomial Theorem | NCERT Solutions for Class 11 Maths

Expand each of the expression in Exercises 1 to 5.
Ex 7.1 Question 1.

$(1-2 x)^5$

Answer.

Using Binomial Theorem,
$
\begin{aligned}
& (1-2 x)^5={ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1(-2 x)+{ }^5 \mathrm{C}_2(-2 x)^2+{ }^5 \mathrm{C}_3(-2 x)^3+{ }^5 \mathrm{C}_4(-2 x)^4+{ }^5 \mathrm{C}_5(-2 x)^5 \\
& =1+5(-2 x)+10(-2 x)^2+10(-2 x)^3+5(-2 x)^4+(-2 x)^5 \\
& =1+5(-2 x)+10\left(4 x^2\right)+10\left(-8 x^3\right)+5\left(16 x^4\right)+\left(-32 x^5\right) \\
& =1-10 x+40 x^2-80 x^3+80 x^4-32 x^5 \text { Ans. }
\end{aligned}
$
Ex 7.1 Question 2.

$\left(\frac{2}{x}-\frac{x}{2}\right)^5$

Answer.

Using Binomial Theorem,
$
\begin{aligned}
& \left(\frac{2}{x}-\frac{x}{2}\right)^5={ }^5 C_0\left(\frac{2}{x}\right)^5+{ }^5 C_1\left(\frac{2}{x}\right)^4\left(\frac{-x}{2}\right)^1+{ }^5 C_2\left(\frac{2}{x}\right)^3\left(\frac{-x}{2}\right)^2+{ }^5 C_3\left(\frac{2}{x}\right)^2\left(\frac{-x}{2}\right)^3+{ }^5 C_4\left(\frac{2}{x}\right)^1 \\
& =\frac{32}{x^5}+5\left(\frac{16}{x^4}\right)\left(\frac{-x}{2}\right)+10\left(\frac{8}{x^3}\right)\left(\frac{x}{4}\right)+10\left(\frac{4}{x^2}\right)\left(\frac{-x^3}{8}\right)+5\left(\frac{2}{x}\right)\left(\frac{x}{16}{ }^4\right)+\left(\frac{-x^5}{32}\right) \\
& =\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5 x+\frac{5}{8} x^3-\frac{x^5}{32}
\end{aligned}
$

Ex 7.1 Question 3.

$(2 x-3)^6$

Answer.

Using Binomial Theorem,

$
\begin{aligned}
& (2 x-3)^6={ }^6 \mathrm{C}_0(2 x)^6+{ }^6 \mathrm{C}_1(2 x)^5(-3)+{ }^6 \mathrm{C}_2(2 x)^4(-3)^2+{ }^6 \mathrm{C}_3(2 x)^3(-3)^3+{ }^6 \mathrm{C}_4(2 x)^2(-3)^4 \\
& +{ }^6 \mathrm{C}_5(2 x)(-3)^5+{ }^6 \mathrm{C}_6(-3)^6 \\
& = \\
& 64 x^6+6\left(32 x^5\right)(-3)+15\left(16 x^4\right)(9)+20\left(8 x^3\right)(-27)+15\left(4 x^2\right)(81)+6(2 x)(-243)+7 \\
& =64 x^6-576 x^5+2160 x^4-4320 x^3+4860 x^2-2916 x+729
\end{aligned}
$
Ex 7.1 Question 4.

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

Answer.

Using Binomial Theorem,
$
\begin{aligned}
& \left(\frac{x}{3}+\frac{1}{x}\right)^5={ }^5 \mathrm{C}_0\left(\frac{x}{3}\right)^5+{ }^5 \mathrm{C}_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)+{ }^5 \mathrm{C}_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2+{ }^5 \mathrm{C}_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3+{ }^5 \mathrm{C}_4\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^4+{ }^5 \mathrm{C}_5\left(\frac{1}{x}\right)^5 \\
& =\frac{x^5}{243}+5 \cdot \frac{x^4}{81} \cdot \frac{1}{x}+10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2}+10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3}+5 \cdot \frac{x}{3} \cdot \frac{1}{x^4}+\frac{1}{x^5} \\
& =\frac{x^5}{243}+\frac{5}{81} x^3+\frac{10}{27} x+\frac{10}{9 x}+\frac{5}{3 x^3}+\frac{1}{x^5}
\end{aligned}
$

Ex 7.1 Question 5.

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

Answer.

Using Binomial Theorem,
$
\begin{aligned}
& \left(x+\frac{1}{x}\right)^6={ }^6 \mathrm{C}_0 x^6+{ }^6 \mathrm{C}_1 x^5\left(\frac{1}{x}\right)+{ }^6 \mathrm{C}_2 x^4\left(\frac{1}{x}\right)^2+{ }^6 \mathrm{C}_3 x^3\left(\frac{1}{x}\right)^3+{ }^6 \mathrm{C}_4 x^2\left(\frac{1}{x}\right)^4+{ }^6 \mathrm{C}_5 x\left(\frac{1}{x}\right)^5+{ }^6 \mathrm{C}_6\left(\frac{1}{x}\right)^6 \\
& =x^6+6 x^5 \cdot \frac{1}{x}+15 x^4 \cdot \frac{1}{x^2}+20 x^3 \cdot \frac{1}{x^3}+15 x^2 \cdot \frac{1}{x^4}+6 x \cdot \frac{1}{x^5}+\frac{1}{x^6} \\
& =x^6+6 x^4+15 x^2+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}
\end{aligned}
$

Using binomial theorem evaluate each of the following:
Ex 7.1 Question 6.

$(96)^3$

Answer.

First we have to express 96 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write $96=100-4$
Therefore $(96)^3=(100-4)^3$
Using Binomial Theorem,
$
\begin{aligned}
& (100-4)^3={ }^3 \mathrm{C}_0(100)^3+{ }^3 \mathrm{C}_1(100)^2(-4)+{ }^3 \mathrm{C}_2(100)(-4)^2+{ }^3 \mathrm{C}_3(-4)^3 \\
& =(100)^3+3.10000(-4)+3.100 .16+(-64) \\
& =1000000-120000+4800-64 \\
& =1004800-120064=884736
\end{aligned}
$

Ex 7.1 Question 7.

$(102)^5$

Answer.

First we have to express 102 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write $102=100+2$
Therefore $(102)^5=(100+2)^5$
Using Binomial Theorem,
$
\begin{aligned}
& (100+2)^5={ }^5 \mathrm{C}_0(100)^5+{ }^5 \mathrm{C}_1(100)^4(2)+{ }^5 \mathrm{C}_2(100)^3(2)^2+{ }^5 \mathrm{C}_3(100)^2(2)^3+{ }^5 \mathrm{C}_4(100)(2)^4+{ }^5 \mathrm{C}_5(2)^5 \\
& =(100)^5+5(100)^4(2)+10(100)^3(2)^2+10(100)^2(2)^3+5.100(2)^4+(2)^5 \\
& =10000000000+1000000000+40000000+800000+8000+32
\end{aligned}
$

Ex 7.1 Question 8.

$(101)^4$

Answer.

First we have to express 101 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write $101=100+1$
Therefore $(101)^4=(100+1)^4$
Using Binomial Theorem,
$
\begin{aligned}
& (100+1)^4={ }^4 C_0(100)^4+{ }^4 C_1(100)^3(1)+{ }^4 C_2(100)^2(1)^2+{ }^4 C_3(100)^1(1)^3+{ }^4 C_4(1)^4 \\
& =(100)^4+4(100)^3+6(100)^2+4(100)+1 \\
& =100000000+4000000+60000+400+1 \\
& =104060401
\end{aligned}
$
Ex 7.1 Question 9.

$(99)^5$

Answer.

First we have to express 99 as the sum or difference of two numbers whose powers are easier to calculate and then use Binomial Theorem

We can write $99=100-1$

Therefore $(99)^5=(100-1)^5$
Using Binomial Theorem,
$
\begin{aligned}
& (100-1)^5={ }^5 \mathrm{C}_0(100)^5+{ }^5 \mathrm{C}_1(100)^4(-1)+{ }^5 \mathrm{C}_2(100)^3(-1)^2+{ }^5 \mathrm{C}_3(100)^2(-1)^3+{ }^5 \mathrm{C}_4(100)(-1)^4+{ }^5 \mathrm{C}_5(-1)^5 \\
& =100^5+5(100)^4(-1)+10(100)^3(1)+10(100)^2(-1)+5(100)(1)+(-1) \\
& =10000000000-500000000+10000000-100000+500-1
\end{aligned}
$

$
=9509900499
$
Ex 7.1 Question 10.

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

Answer.

We have $1.1=1+0.1$
$
\therefore(1.1)^{10000}=(1+0.1)^{10000}
$

Using Binomial Theorem,
$
\begin{aligned}
& (1+0.1)^{10000}=1+{ }^{10000} \mathrm{C}_1(.1)+{ }^{10000} \mathrm{C}_2(0.1)^2+{ }^{10000} \mathrm{C}_3(0.1)^3+\ldots \\
& =1+10000 \text { (0.1) + other positive numbers } \\
& =1+1000 \text { + other positive numbers }
\end{aligned}
$
which is greater than 1000
$
\therefore(1.1)^{10000}>1000
$
Ex 7.1 Question 11.

Find $(a+b)^4-(a-b)^4$. Hence evaluate: $(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4$.

Answer.

Given: $(a+b)^4-(a-b)^4$
Using Binomial Theorem,

$\begin{aligned}
& (a+b)^4-(a-b)^4=\left[{ }^4 \mathrm{C}_0 a^4+{ }^4 \mathrm{C}_1 a^3 b+{ }^4 \mathrm{C}_2 a^2 b^2+{ }^4 \mathrm{C}_3 a b^3+{ }^4 \mathrm{C}_4 b^4\right] \\
& -\left[{ }^4 \mathrm{C}_0 a^4+{ }^4 \mathrm{C}_1 a^3(-b)+{ }^4 \mathrm{C}_2 a^2(-b)^2+{ }^4 \mathrm{C}_3 a(-b)^3+{ }^4 \mathrm{C}_4(-b)^4\right] \\
& =a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4-\left[a^4-4 a^3 b+6 a^2 b^2-4 a b^3+b^4\right] \\
& =a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4-a^4+4 a^3 b-6 a^2 b^2+4 a b^3-b^4 \\
& =8 a^3 b+8 a b^3=8 a b\left(a^2+b^2\right)
\end{aligned}$

Putting $a=\sqrt{3}$ and $b=\sqrt{2}$
$
\begin{aligned}
& (\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4=8 \sqrt{3} \cdot \sqrt{2}\left[(\sqrt{3})^2+(\sqrt{2})^2\right] \\
& =8 \sqrt{6}[3+2]=40 \sqrt{6} \text { Ans. }
\end{aligned}
$
Ex 7.1 Question 12.

Find $(x+1)^6+(x-1)^6$. Hence or otherwise evaluate $(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$.

Answer.

Given: $(x+1)^6+(x-1)^6$
Using Binomial Theorem,
$
\begin{aligned}
& (x+1)^6+(x-1)^6=\left[{ }^6 \mathrm{C}_0 x^6+{ }^6 \mathrm{C}_1 x^5(1)+{ }^6 \mathrm{C}_2 x^4(1)^2+{ }^6 \mathrm{C}_3 x^3(1)^3+{ }^6 \mathrm{C}_4 x^2(1)^4+{ }^6 \mathrm{C}_5 x(1)^5+{ }^6 \mathrm{C}_6(1)^6\right] \\
& +\left[{ }^6 \mathrm{C}_0 x^6+{ }^6 \mathrm{C}_1 x^5(-1)+{ }^6 \mathrm{C}_2 x^4(-1)^2+{ }^6 \mathrm{C}_3 x^3(-1)^3+{ }^6 \mathrm{C}_4 x^2(-1)^4+{ }^6 \mathrm{C}_5 x(-1)^5+{ }^6 \mathrm{C}_6(-1)^6\right] \\
& =x^6+6 x^5+15 x^4+20 x^3+15 x^2+6 x+1+x^6-6 x^5+15 x^4-20 x^3+15 x^2-6 x+1 \\
& =2 x^6+30 x^4+30 x^2+2=2\left[x^6+15 x^4+15 x^2+1\right]
\end{aligned}
$

Putting $x=\sqrt{2}$

$
\begin{aligned}
& (\sqrt{2}+1)^6+(\sqrt{2}-1)^6=2\left[(\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1\right] \\
& =2[8+15 \times 4+15 \times 2+1] \\
& =2[8+60+30+1]=2 \times 99=198
\end{aligned}
$
Ex 7.1 Question 13.

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

Answer.

We know that $\mathrm{b}$ is divisible by $\mathrm{a}$ ( or a divides $\mathrm{b}$ ) $\Rightarrow b=a k$, $\mathrm{k}$ is an integer Here we have to show that 64 divides $9^{n+1}-8 n-9$

$\Rightarrow 9^{n+1}-8 n-9=64 k, \mathrm{k}$ is an integer
We have $9^{n+1}=(1+8)^{n+1}$
Using Binomial Theorem,we have
$
(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots{ }^n C_n x^n
$

$\begin{aligned}
& \therefore 9^{n+1}=(1+8)^{n+1}={ }^{n+1} \mathrm{C}_0+{ }^{n+1} \mathrm{C}_1(8)+{ }^{n+1} \mathrm{C}_2(8)^2+{ }^{n+1} \mathrm{C}_3(8)^3+\ldots \ldots+{ }^{n+1} \mathrm{C}_{n+1}(8)^{n+1} \\
& =1+(n+1) \times 8+{ }^{n+1} \mathrm{C}_2(8)^2+{ }^{n+1} \mathrm{C}_3(8)^3+\ldots \ldots+{ }^{n+1} \mathrm{C}_{n+1}(8)^{n+1} \\
& =1+8 n+8+{ }^{n+1} \mathrm{C}_2(8)^2+{ }^{n+1} \mathrm{C}_3(8)^3+\ldots \ldots+{ }^{n+1} \mathrm{C}_{n+1}(8)^{n+1} \\
& \Rightarrow 9^{n+1}-8 n-9={ }^{n+1} \mathrm{C}_2(8)^2+{ }^{n+1} \mathrm{C}_3(8)^3+\ldots \ldots . .+{ }^{n+1} \mathrm{C}_{n+1}(8)^{n+1}
\end{aligned}$

$
9^{n+1}-8 n-9=64\left[{ }^{n+1} C_2+{ }^{n+1} C_3(8)+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+{ }^{n+1} C_{n+1}(8)^{n-1}\right]
$
$\Rightarrow 9^{n+1}-8 n-9=64 k$, where $k={ }^{n+1} C_2+{ }^{n+1} C_3(8)+\ldots \ldots \ldots .+{ }^{n+1} C_{n+1}(8)^{n-1}$ is an integer
which shows that $9^{n+1}-8 n-9$ is divisible by 64 .
Ex 7.1 Question 14.

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

Answer.

L.H.S. $=\sum_{r=0}^n 3^r{ }^n \mathrm{C}_r=3^0{ }^n \mathrm{C}_0+3^{1 n} \mathrm{C}_1+3^2{ }^n \mathrm{C}_2+\ldots \ldots \ldots . .+3^n{ }^n \mathrm{C}_n$
But we have
$
\begin{aligned}
& (1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+{ }^n C_n x^n \\
& \therefore \sum_{r=0}^n 3{ }^{r n} C_r=(1+3)^n=4^n
\end{aligned}
$

Hence proved