If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then $C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$ equals

  • A
    $\frac{(2n)!}{(n+1)!(n+2)!}$
  • B
    $\frac{(2n)!}{(n-2)!(n+2)!}$
  • C
    $\frac{(2n)!}{n!(n+2)!}$
  • D
    $\frac{(2n)!}{(n-1)!(n+2)!}$

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The value of $\frac{C_1}{C_0} + 2 \cdot \frac{C_2}{C_1} + 3 \cdot \frac{C_3}{C_2} + \dots + n \cdot \frac{C_n}{C_{n-1}}$ is equal to

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Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

If for $3 \leq r \leq 30$, $\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}$, then $m$ equals:

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

$\binom{n}{n-r} + \binom{n}{r+1}$,whenever $0 \le r \le n-1$,is equal to

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