Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)^99.

Question

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+x)^{99}$. Let $a$ be the middle term in the expansion of $\left(2+\dfrac{1}{\sqrt2}\right)^{200}$. If $\dfrac{{}^{200}C_{99}\,K}{a}=2^{\ell}\cdot\dfrac{m}{n}$, where $m$ and $n$ are odd numbers, then the ordered pair $(\ell,n)$ is equal to:

Accepted Answer

$K={}^{99}C_1+{}^{99}C_3+\cdots+{}^{99}C_{99}=2^{98}$.

Middle term of $\left(2+\frac{1}{\sqrt2}\right)^{200}$ is $T_{101}=\binom{200}{100}2^{100}\left(\frac{1}{\sqrt2}\right)^{100}=\binom{200}{100}2^{50}$.

$$\frac{{}^{200}C_{99}\times2^{98}}{{}^{200}C_{100}\times2^{50}}=\frac{100}{101}\times2^{48}$$

So $\dfrac{25\cdot2^{50}}{101}=\dfrac{m}{n}2^\ell$ --- matching powers gives $(\ell,n)=(50,101)$.

Solutions · 1