If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ are Binomial Coefficient in the expansion of ${(1 + x)}^{n}$ then value of $C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1}$ equals

\frac{3^{n + 1} - 1}{n + 1}

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

\frac{3^{n + 1} + 1}{n + 1}

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

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

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

Open in App

Solution

The correct option is C $\frac{2^{n + 1} - 1}{n + 1}$
$(1 + x)^{n} = 1 +^{n} C_{1} x +^{n} C_{2} x^{2} + . . .^{n} C_{n} x^{n}$
$\int_{0}^{1} (1 + x)^{n} . d x = 1 + \frac{^{n} C_{1}}{2} + \frac{^{n} C_{3}}{3} + \frac{^{n} C_{4}}{2} . . . + \frac{^{n} C_{n}}{n + 1}$
Hence
$\int_{0}^{1} (1 + x)^{n} . d x$
$[\frac{(1 + x)^{n + 1}}{n + 1}]_{0}^{1}$
$= \frac{2^{n + 1} - 1}{n + 1}$

Suggest Corrections

Similar questions

C_{0}, C_{1}, C_{2}, . . . . . . . . . . . C_{n}

(1 + x)^{n} .

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . . (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1})

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

{(1 + x)}^{n}

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

c_{0}, c_{1}, c_{2}, . . . . c_{n}

(1 + x)^{n}

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

c_{0}, c_{1}, c_{2}

(1 + x)^{n}

c_{1} + c_{1} c_{2} + c_{2} c_{3} + . . . . . . . c_{n - 1} c_{n} = \frac{(2 n)!}{(n + 1)! (n - 1)!}

C_{0}, C_{1}, C_{2} . . . . . . . . . . ., C_{n}

(1 + x)^{n}, n

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . + C_{n - 1})

Join BYJU'S Learning Program