If (1+x)^n = C_0 + C_1 x + C_2 x^2 + ldots + | vedclass

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(C) Given the binomial expansion: $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$. Integrate both sides with respect to $x$ from $0$ to $1$: $\in

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$\frac{2^{n+1}-1}{n+1}$

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(C) Given the binomial expansion: $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$. Integrate both sides with respect to $x$ from $0$ to $1$: $\int_{0}^{1} (1+x)^n dx = \int_{0}^{1} (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) dx$. Evaluating the integral on the left side: $\left[ \frac{(1+x)^{n+1}}{n+1} \right]_{0}^{1} = \frac{(1+1)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{2^{n+1}-1}{n+1}$. Evaluating the integral on the right side: $\left[ C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{3} + \ldots + \frac{C_n x^{n+1}}{n+1} \right]_{0}^{1} = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1}$. Equating both sides,we get: $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} = \frac{2^{n+1}-1}{n+1}$.

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$ for $n \in N$,then $C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} =$

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