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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

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

S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t}),

l,

l = \infty \sum n = 1 (1 - S_{n})

n ϵ N

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

2 n \sum r = 0 a_{2 r}

n \sum r = 1 t_{r} = \frac{n (n + 1) (n + 2) (n + 3)}{8}

n \sum r = 1 \frac{1}{t_{r}}

Join BYJU'S Learning Program