First slide

Binomial theorem for positive integral Index

Question

$If (1 + x)^{n} = C + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}, then the value of C_{0} + \frac{1}{2} C_{1} + \frac{1}{3} C_{2} + \dots + \frac{1}{(n + 1)} C_{n} is$

Difficult

A

$\frac{2^{n - 1}}{(n + 1)}$
B

$\frac{2^{n + 1}}{(n + 1)}$
C

$\frac{2^{n - 1} 1}{(n + 1)}$
D

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

Solution

$∴ (1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n} \int_{0}^{1} (1 + x)^{n} d x = \int_{0}^{1} (C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}) d x \frac{2^{n + 1}}{n + 1} - \frac{1}{n + 1} = C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1}$

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