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The sum to (n+1) terms of the following series $$ \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots . . $$ is

Option: 1

\frac{1}{n+1}


Option: 2

\frac{1}{n+2}


Option: 3

\frac{1}{n(n+1)}


Option: 4

None of these


Answers (1)

best_answer

(1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots \ldotsx(1-x)^n=C_0 x-C_1 x^2+C_2 x^3-C_3 x^4+\ldots \ldots

\int_0^1 x(1-x)^n d x=C_0\left[\frac{x^2}{2}\right]_0^1-C_1\left[\frac{x^3}{3}\right]_0^1+C_2\left[\frac{x^4}{4}\right]_0^1-\ldots \ldots .

The integral on L.H.S. of (i)=\int_1^0(1-t) t^n(-d t) by putting 1-x=t 

\int_0^1\left(t^n-t^{n+1}\right) d t=\frac{1}{n+1}-\frac{1}{n+2}

Whereas the integral on the R.H.S. of (i)

\begin{aligned} & =C_0\left[\frac{1}{2}\right]-C_1\left[\frac{1}{3}\right]+\frac{C_2}{4}-\ldots \ldots . . \\ & \frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)} \end{aligned}

Trick : Put n=1 in given series \frac{{ }^1 C_0}{2}-\frac{{ }^1 C_1}{3}=\frac{1}{6} Which is given by option (d).

 

Posted by

Kuldeep Maurya

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