Let K be the sum of the coefficients of the o | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

In the expansion of

$(1+ x )^{99}= C _0+ C _1 x + C _2 x ^2+\ldots+ C _{99} x ^{99}$

$K = C _1+ C _3+\ldots . .+ C _{99}=2^{98}$

$a$ Middle i

Vedclass · Accepted Answer

$(50,101)$

Vedclass · Answer

In the expansion of

$(1+ x )^{99}= C _0+ C _1 x + C _2 x ^2+\ldots+ C _{99} x ^{99}$

$K = C _1+ C _3+\ldots . .+ C _{99}=2^{98}$

$a$ Middle in the expansion of $\left(2+\frac{1}{\sqrt{2}}ight)^{200}$

$\frac{T_{200}}{2}+1 ={ }^{200} C _{100}(2)^{100}\left(\frac{1}{\sqrt{2}}ight)^{100}$

$={ }^{200} C _{100} \cdot 2^{50}$

So, $\frac{{ }^{200} C _{99} 	imes 2^{98}}{{ }^{900} C _{100} 	imes 2^{50}}=\frac{100}{101} 	imes 2^{48}$

So, $\frac{25}{101} 	imes 2^{50}=\frac{ m }{ n } 2^{\prime}$

$	herefore m , n$ are odd so

$(\ell, n )$ become $(50,101)$ Ans.

Similar Questions

In the expansion of ${(x + a)^n}$, the sum of odd terms is $P$ and sum of even terms is $Q$, then the value of $({P^2} - {Q^2})$ will be

If $n$ is an integer greater than $1$, then $a{ - ^n}{C_1}(a - 1){ + ^n}{C_2}(a - 2) + .... + {( - 1)^n}(a - n) = $

Value $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r}{}^{40}{C_{15}}{}^{20}{C_r} - {}^{35}{C_{15}}{}^{15}{C_r}{}^{25}{C_r}} \right)} $ is-

If the sum of the coefficients in the expansion of ${(\alpha {x^2} - 2x + 1)^{35}}$ is equal to the sum of the coefficients in the expansion of ${(x - \alpha y)^{35}}$, then $\alpha $=

The value of ${\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}\left( { - 1} \right)}}{{{2^{2r + 1}}}}} ^r}$ is