ધારો કે (1+x)^{99}ના વિસ્તરણમાં xની અયુગ | vedclass

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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.

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