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  •  If <img alt="\begin{array}{l}{ \mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots, \mathrm{C}_{n} }\end{array}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cbegin%7Barray%

 If \begin{array}{l}{ \mathrm{C}_{0}, \mathrm{C}_{1}, \mathrm{C}_{2}, \ldots, \mathrm{C}_{n} }\end{array} be binomial coefficients in the expansion of (1+x)^nFind the value of C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\ldots+(-1)^{n} \frac{C_{n}}{n+1}.

 

Option: 1

\frac{n}{n-1}


Option: 2

\frac{n}{n+1}


Option: 3

\frac{1}{n-1}


Option: 4

\frac{1}{n+1}


Answers (1)

best_answer

(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{n} x^{n}

Integrating within limits -1 to 0, then we get,

\int_{-1}^{0}(1+x)^{n} d x=\int_{-1}^{0}\left(C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{n} x^{n}\right) d x

\begin{array}{l}{\Rightarrow\left[\frac{(1+x)^{n+1}}{n+1}\right]_{-1}^{0}=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\ldots+\frac{C_{n} x^{n+1}}{n+1}\right]_{-1}^{0}} \\\\ {\Rightarrow \frac{1-0}{n+1}=0-\left(-C_{0}+\frac{C_{1}}{2}-\frac{C_{2}}{3}+\ldots+(-1)^{n+1} \frac{C_{n}}{n+1}\right)} \\\\ {\Rightarrow \frac{1}{n+1}=C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\ldots+(-1)^{n+2} \frac{C_{n}}{n+1}}\end{array}

\begin{array}{l}{\Rightarrow \frac{1}{n+1}=C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\ldots+(-1)^{n} \frac{C_{n}}{n+1}} \\\\\because \left[(-1)^{n+2}=(-1)^{n}(-1)^{2}=(-1)^{n}\right]\\\\ {\text { Hence, } C_{0}-\frac{C_{1}}{2}+\frac{C_{2}}{3}-\ldots+(-1)^{n} \frac{C_{n}}{n+1}=\frac{1}{n+1}}\end{array}

option D is correct.

Posted by

HARSH KANKARIA

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