If $(1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$, then $\sum_{k=0}^7 (-1)^k a_{2k}$ is equal to

  • A
    $2^5$
  • B
    $4^5$
  • C
    $0$
  • D
    $4^4$

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Let $C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$,and $C_{1}+3 \cdot 2 C_{2}+5 \cdot 3 C_{3}+\ldots$ (up to $10$ terms) $= \frac{\alpha \times 2^{11}}{2^{\beta}-1} \left( C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\ldots \right.$ (up to $10$ terms) $)$,then the value of $\alpha+\beta$ is equal to:

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

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If $C_r$ denotes the binomial coefficient ${ }^{n} C_r$,then $(-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2$ is equal to

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