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  • <img alt="\text { If }(1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n \text {, }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%20%7B%20If%20%7D%281-x%29%5En%3DC_0-C_1%20x&

\text { If }(1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n C_n x^n \text {, }\text { then } \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\ldots+(-1)^n \frac{C_n}{n+2}=

Option: 1

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


Option: 2

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


Option: 3

\frac{1}{n^2}


Option: 4

None of these 


Answers (1)

best_answer

 Now,

(1-x)^n=C_0-C_1 x+C_2 x^2-C_3 x^3+\ldots .+(-1)^n n C_n x^n
Multiplying both sides of above by x,

x(1-x)^n=C_0 x-C_1 x^2+C_2 x^3 \ldots . .+(-1)^n C_n x^{n+1}
Integrating both sides, from 0 to 1.

\begin{array}{r} \int_0^1 x(1-x)^n d x=\int_0^1\left\{C_0 x-C_1 \cdot x^2+\ldots+(-1)^n \cdot C_n \cdot x^{n+1}\right\} d x \\ \Rightarrow \int_0^1(1-x) x^n d x=\int_0^1\left\{C_0 x-C_1 \cdot x^2+\ldots+(-1)^n \cdot C_n \cdot x^{n+1}\right\} d x \\ {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \end{array}\begin{array}{r} \Rightarrow\left(\frac{x^{n+1}}{n+1}\right)_0^1-\left(\frac{x^{n+2}}{n+2}\right)_0^1=C_0\left(\frac{x^2}{2}\right)_0^1-C_1 \cdot\left(\frac{x^3}{3}\right)_0^1+\ldots \\ +(-1)^n \cdot C_n\left(\frac{x^{n+2}}{n+2}\right)_0^1 \end{array}

On putting the limits we get the option b

Posted by

Sanket Gandhi

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