MCQ
$\left\{C_1+2 C_2+3 C_3+\ldots+n C_n\right\}=?$
  • A
    $( n -1) \cdot 2^{ n }$
  • B
    $n \cdot 2^n$
  • C
    $( n +1) \cdot 2^n$
  • D
    $n \cdot 2^{n-1}$

Answer

(d) $n \cdot 2^{ n -1}$
Explanation: $C _1+2 C _2+3 C _3+\ldots+n C _{ n }= n +2 \cdot \frac{ n ( n -1)}{2}+3 \cdot \frac{n(n-1)(n-2)}{3!}+\ldots+n$
$\begin{array}{l}= n \cdot\left[1+( n -1) \frac{( n -1)( n -2)}{2!}+\ldots+1\right] \\ = n \cdot\left[{ }^{( n -1)} C _0+{ }^{( n -1)} C _1+{ }^{( n -1)} C _2+\ldots+{ }^{( n -1)} C _{ n -1}\right] \\ = n \cdot(1+1)^{ n -1}= n \cdot 2 ^{ n -1}\end{array}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free