Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  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 \frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+\ldots

Option: 1

\frac{2^{n}+1}{n+1}


Option: 2

\frac{2^{n}-1}{n+1}


Option: 3

\frac{2^{n}-1}{n-1}


Option: 4

\frac{2^{n}+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 between limits 0 and 1, we get,

\begin{array}{l} \\ {\int_{0}^{1}(1+x)^{n} d x=\int_{0}^{1} \left (C_{0} d x+\int_{0}^{1} C_{1} x d x+\int_{0}^{1} C_{2} x^{2} d x+\ldots+\int_{0}^{1} C_{n} x^{n} \right ) d x} \\\\ {\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{1}=\left.\frac{C_{0} x}{1}\right|_{0} ^{1}+C_{1}\left.\frac{x^{2}}{2}\right|_{0} ^{1}+C_{2} \cdot\left.\frac{x^{3}}{3}\right|_{0} ^{1}+\ldots+C_{n} \cdot\left.\frac{x^{n+1}}{n+1}\right|_{0} ^{1}}\end{array}

\begin{array}{l}{\Rightarrow \quad \frac{2^{n+1}}{n+1}-\frac{1}{n+1}=C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\ldots+\frac{C_{n}}{n+1}} \\ \\{\Rightarrow \quad C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\ldots+\frac{C_{n}}{n+1}=\frac{2^{n+1}-1}{n+1}.........(1)}\end{array}

 

Now, 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}}{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}.......(2)}\end{array}

 

On subtracting equation 2 from eq 1.....

2\left(\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+\ldots\right)=\frac{2^{n+1}-2}{n+1}

\frac{C_{1}}{2}+\frac{C_{3}}{4}+\frac{C_{5}}{6}+\ldots=\frac{2^{n}-1}{n+1}

Option B is correct

 

Posted by

seema garhwal

View full answer

Similar Questions

Latest Question