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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}+..... 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 . .

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

\Rightarrow \int_0^1 x(1-x)^n d x=\int_0^1\left(C_0 x-C_1 x^2+C_2 x^3 \ldots .\right) d x …...(i)

The integral on the LHS

=\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 RHS of (i)

=\left[\frac{C_0 x^2}{2}-\frac{C_1 x^3}{3}+\frac{C_2 x^4}{4}-\ldots\right]=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\ldots

\therefore \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\ldots  to (n-1) terms

=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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