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

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

Example 1

Expand $\left(x^2+\frac{3}{x}\right)^4, x \neq 0$
Solution

By using binomial theorem, we have
$
\begin{aligned}
x^2+\frac{3}{x} & ={ }^4 \mathrm{C}_0\left(x^2\right)^4+{ }^4 \mathrm{C}_1\left(x^2\right)^3\left(\frac{3}{x}\right)+{ }^4 \mathrm{C}_2\left(x^2\right)^2\left(\frac{3}{x}\right)^2+{ }^4 \mathrm{C}_3\left(x^2\right)\left(\frac{3}{x}\right)^3+{ }^4 \mathrm{C}_4\left(\frac{3}{x}\right)^4 \\
& =x^8+4 \cdot x^6 \cdot \frac{3}{x}+6 \cdot x^4 \cdot \frac{9}{x^2}+4 \cdot x^2 \cdot \frac{27}{x^3}+\frac{81}{x^4} \\
& =x^8+12 x^5+54 x^2+\frac{108}{x}+\frac{81}{x^4} .
\end{aligned}
$

Example 2

Compute $(98)^5$.
Solution

We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write $98=100-2$
Therefore, $(98)^5=(100-2)^5$
$
\begin{aligned}
= & { }^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 \\
= & 10000000000-5 \times 100000000 \times 2+10 \times 1000000 \times 4-10 \times 10000 \\
& \times 8+5 \times 100 \times 16-32 \\
= & 10040008000-1000800032=9039207968 .
\end{aligned}
$

Solution

Splitting 1.01 and using binomial theorem to write the first few terms we have

$
\begin{aligned}
&(1.01)^{1000000}=(1+0.01)^{1000000} \\
&={ }^{1000000} \mathrm{C}_0+{ }^{1000000} \mathrm{C}_1(0.01)+\text { other positive terms } \\
&=1+1000000 \times 0.01+\text { other positive terms } \\
&= 1+10000+\text { other positive terms } \\
&>10000
\end{aligned}
$

Hence
$
(1.01)^{1000000}>10000
$

Example 4

Using binomial theorem, prove that $6^n-5 n$ always leaves remainder 1 when divided by 25 .

Solution

For two numbers $a$ and $b$ if we can find numbers $q$ and $r$ such that $a=b q+r$, then we say that $b$ divides $a$ with $q$ as quotient and $r$ as remainder. Thus, in order to show that $6^n-5 n$ leaves remainder 1 when divided by 25 , we prove that $6^n-5 n=25 k+1$, where $k$ is some natural number.

We have
$
(1+a)^n={ }^n \mathrm{C}_0+{ }^n \mathrm{C}_1 a+{ }^n \mathrm{C}_2 a^2+\ldots+{ }^n \mathrm{C}_n a^n
$

For $a=5$, we get
$
(1+5)^n={ }^n \mathrm{C}_0+{ }^n \mathrm{C}_1 5+{ }^n \mathrm{C}_2 5^2+\ldots+{ }^n \mathrm{C}_n 5^n
$
i.e.
$
(6)^n=1+5 n+5^2 . \cdot \mathrm{C}_2+5^3 \cdot{ }^n \mathrm{C}_3+\ldots+5^n
$
i.e. $\quad 6^n-5 n=1+5^2\left({ }^n \mathrm{C}_2+{ }^n \mathrm{C}_3 5+\ldots+5^{n-2}\right)$
or
$
6^n-5 n=1+25\left({ }^n \mathrm{C}_2+5 \cdot{ }^n \mathrm{C}_3+\ldots+5^{n-2}\right)
$
or
$6^n-5 n=25 k+1 \quad$ where $k={ }^n \mathrm{C}_2+5 \cdot{ }^n \mathrm{C}_3+\ldots+5^{n-2}$