MCQ
$2{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + .... + \frac{{{2^{11}}}}{{11}}{C_{10}}$ = . . .
- ✓$\frac{{{3^{11}} - 1}}{{11}}$
- B$\frac{{{2^{11}} - 1}}{{11}}$
- C$\frac{{{{11}^3} - 1}}{{11}}$
- D$\frac{{{{11}^2} - 1}}{{11}}$
Integrating both sides from $0$ to $2$, we get
$\frac{{{3^{11}} - 1}}{{11}} = 2{C_0} + \frac{{{2^2}}}{2}{C_1} + \frac{{{2^3}}}{3}{C_2} + .... + \frac{{{2^{11}}}}{{11}}{C_{10}}$.
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then the value of $6 \beta$ is $\qquad$
$[A]$ $2 a, 4,1$ $[B]$ $2 a, 8,1$ $[C]$ $a, 4,1$ $[D]$ $a, 4,2$