### Binomial Theorem Question 11

#### Question 11 - 29 January - Shift 2

Let $K$ be the sum of the coefticients of the odd powers of $x$ in the expansion of $(1+x)^{99}$. Let a be

the middle term in the expansion of $(2+\frac{1}{\sqrt{2}})^{200}$. If $\frac{{ }^{200} C _{99} K}{a}=\frac{2^{\ell} m}{n}$, where $m$ and $n$ are odd numbers, then the ordered pair $(\ell, n)$ is equal to :

(1) $(50,51)$

(2) $(51,99)$

(3) $(50,101)$

(4) $(51,101)$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Bionomial Theorem, Middle term in the Expansion when n is even, General term of Bionomial Cofficient

In the expansion of

$(1+x)^{99}=C_0+C_1 x+C_2 x^{2}+\ldots .+C _{99} x^{99}$

$K=C_1+C_3+\ldots .+C _{99}=2^{98}$

$a \Rightarrow$ Middle in the expansion of $(2+\frac{1}{\sqrt{2}})^{200}$

$T _{\frac{200}{2}+1}={ }^{200} C _{100}(2)^{100}(\frac{1}{\sqrt{2}})^{100}$

$ ={ }^{200} C _{100} \cdot 2^{50} $

So, $\frac{{ }^{200} C _{99} \times 2^{98}}{{ }^{200} C _{100} \times 2^{50}}=\frac{100}{101} \times 2^{48}$

So, $\frac{25}{101} \times 2^{50}=\frac{m}{n} 2^{\ell}$

$\therefore \quad m, n$ are odd so $(\ell, n)$ become $(50,101)$ Ans.