Q63 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$ for $n \in N$, then $C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots+\frac{C_n}{n+1}=$
- A$\frac{2^n-1}{n+1}$
- Correct answerB$\frac{2^n-1}{n}$
- C$\frac{2^{n+1}-1}{n+1}$
- D$\frac{2^{n+1}-1}{n}$
Solution
$$
(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots . .+C_n x^n
$$
Integrate both sides wrt $x$
$$
\begin{aligned}
& \int(1+x)^n=C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots .+\frac{C_n x^{n+1}}{n+1}+C \\
& \Rightarrow \frac{(1+x)^{n+1}}{n+1}=C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots .+\frac{C_n x^{n+1}}{n+1}+C
\end{aligned}
$$
Put $x=0 \Rightarrow \frac{1}{n+1}=C$
Put $x=1$
$$
\begin{aligned}
& \frac{2^{n+1}}{n+1}-\frac{1}{n+1}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1} \\
& \therefore \quad C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1} .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.