Q123 / 200 · TS EAMCET 2001 · Mathematics

If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$, then $C_0+2 C_1+3 C_2+\ldots+(n+1) C_n$ is equal to

  • Correct answer
    A
    $2^n+n \cdot 2^{n-1}$
  • B
    $2^{n-1}+n \cdot 2^n$
  • C
    $2^n+(n+1) 2^n$
  • D
    $2^{n-1}+(n-1) 2^n$

Solution

Let $S=C_0+2 C_1+3 C_2+\ldots+(n+1) C_n \ldots$ (i) $\begin{aligned} & S=C_n+2 C_{n-1}+3 C_{n-2}+\ldots+(n+1) C_0 \\ & \text { or } S=(n+1) C_0+n C_1+\ldots+C_n\end{aligned}$ On adding Eqs. (i) and (ii), we have $\begin{aligned} 2 S & =C_0[1+n+1]+C_1[2+n]+\ldots+C_n[n+1+1] \\ 2 S & =(n+2)\left(C_0+C_1+\ldots+C_n\right) \\ S & =\frac{(n+2)}{2} 2^n=(n+2) 2^{n-1}=n 2^{n-1}+2^n\end{aligned}$

Hence, the correct answer is option: 'A'.

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