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 $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :
$(50,51)$
$(51,99)$
$(50,101)$
$(51,101)$
$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $
If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $\left(2 x^{3}+\frac{3}{x}\right)^{10}$ is $5^{10}-\beta \cdot 3^{9}$, then $\beta$ is equal to
The sum of the coefficients in the expansion of ${(1 + x - 3{x^2})^{3148}}$ is
If $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ then $n$ is equal to
If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =