श्रेणी $1 + \frac{1}{2} {}^{n}C_{1} + \frac{1}{3} {}^{n}C_{2} + \dots + \frac{1}{n+1} {}^{n}C_{n}$ का योग किसके बराबर है?

  • A
    $\frac{2^{n+1}-1}{n+1}$
  • B
    $\frac{3(2^{n}-1)}{2n}$
  • C
    $\frac{2^{n}+1}{n+1}$
  • D
    $\frac{2^{n}+1}{2n}$

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Similar Questions

मान लीजिए कि $\binom{n}{k}$ का अर्थ ${}^{n}C_{k}$ है और $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{यदि } 0 \leq k \leq n \\ 0, & \text{अन्यथा} \end{cases}$ है। यदि $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ और $A_{4}-A_{3}=190p$ है,तो $p$ का मान ज्ञात कीजिए:

${ }^{34}C_{10} + 3 \cdot { }^{34}C_{9} + 3 \cdot { }^{34}C_{8} + { }^{34}C_{7} = $

मान लीजिए $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,और $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
कथन $(A) : S_3 = 55 \times 2^9$
कारण $(R) : S_1 = 90 \times 2^8$ और $S_2 = 10 \times 2^8$

यदि $C_j = {}^{n}C_j$ है,तो $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

यदि $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$ है, तो $a_2+a_4+a_6+\ldots+a_{12}$ का मान ज्ञात कीजिए।

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