The correct option is B 2n+1 1n+11+xn nbsp;= nbsp;n0 nbsp;+n1x nbsp;+.... nbsp;+ nbsp;nnxnIntegrating nbsp;both nbsp;sides nbsp;from nbsp;0 ...

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Integration to Solve Modified Sum of Binomial Coefficients

∑r =0r=nnrr+1

Question

$r = n \sum r = 0 \frac{(\begin{matrix} n r \end{matrix})}{r + 1}$

2^{n}

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

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n

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n^{n}

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Solution

The correct option is B $\frac{2^{n + 1} - 1}{n + 1}$
${(1 + x)}^{n} = (\begin{matrix} n 0 \end{matrix}) + (\begin{matrix} n 1 \end{matrix}) x + . . . . + (\begin{matrix} n n \end{matrix}) x^{n} Integrating both sides from 0 to 1 \int_{0}^{1} {(1 + x)}^{n} dx = \int_{0}^{1} (\begin{matrix} n 0 \end{matrix}) dx + \int_{0}^{1} (\begin{matrix} n 1 \end{matrix}) x dx + . . . . + \int_{0}^{1} (\begin{matrix} n n \end{matrix}) x^{n} dx \frac{2^{n + 1} - 1}{n + 1} = \frac{(\begin{matrix} n 0 \end{matrix})}{1} + \frac{(\begin{matrix} n 1 \end{matrix})}{2} + . . . . + \frac{(\begin{matrix} n n \end{matrix})}{n + 1} \frac{2^{n + 1} - 1}{n + 1} = r = n \sum r = 0 \frac{(\begin{matrix} n r \end{matrix})}{r + 1}$

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Similar questions

r = n \sum r = 0 \frac{(\begin{matrix} n r \end{matrix})}{r + 1}

$\sum_{r = 1}^{n} (\sum_{k = 0}^{r - 1}^{n} C_{r}^{r} C_{k} 2^{k})$ is equal to

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(1 + 4 x + 4 x^{2})^{n} = r = 2 n \sum r = 0 a_{r} x^{r}

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