જો $\sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2 = \frac{\alpha 60 !}{(30 !)^2}$ હોય,તો $\alpha$ ની કિંમત શોધો.

  • A
    $30$
  • B
    $60$
  • C
    $15$
  • D
    $10$

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Similar Questions

જો $P_{n}$ એ $(1+x)^{n}$ ના વિસ્તરણમાં દ્વિપદી સહગુણકોનો ગુણાકાર દર્શાવે,તો $\frac{P_{n+1}}{P_n}=$

જો $n \in N$ માટે $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$ હોય,તો $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} =$

ધારો કે $\binom{n}{k}$ એ ${}^{n}C_{k}$ દર્શાવે છે અને $\left[\begin{array}{c} n \\ k \end{array}\right]=\begin{cases} \binom{n}{k}, & \text{જો } 0 \leq k \leq n \\ 0, & \text{અન્યથા} \end{cases}$. જો $A_{k}=\sum_{i=0}^{9}\binom{9}{i}\left[\begin{array}{c} 12 \\ 12-k+i \end{array}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\begin{array}{c} 13 \\ 13-k+i \end{array}\right]$ અને $A_{4}-A_{3}=190p$ હોય,તો $p$ ની કિંમત શોધો:

જો ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$ હોય,તો $n=$

$\binom{10}{1} + \binom{10}{2} + \binom{11}{3} + \binom{12}{4} + \binom{13}{5} = \dots$

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