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 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 3 C_{0}+3^{2} \frac{C_{1}}{2}+\frac{3^{3} C_{2}}{3}+\frac{3^{4} C_{3}}{4}+\ldots+\frac{3^{n+1} C_{n}}{n+1}.

Option: 1

\frac{4^{n}-1}{n+1}


Option: 2

\frac{4^{n+1}+1}{n+1}


Option: 3

\frac{4^{n+1}-1}{n+1}


Option: 4

\frac{4^{n}-1}{n+1}


Answers (1)

best_answer

 

 

Use of Integration in Binomial - Part 2 -

Some Binomial Series using Integration

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(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\ldots+C_{n} x^{n}

Integrating within limits 0 to 3, then we get,

\begin{array}{l}{\int_{0}^{3}(1+x)^{n} d x=\int_{0}^{3}\left(C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\ldots+C_{n} x^{n}\right) d x} \\\\ {\Rightarrow\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{3}=\left[C_{0} x+\frac{C_{1} x^{2}}{2}+\frac{C_{2} x^{3}}{3}+\frac{C_{3} x^{4}}{4}+..\ldots+\frac{C_{n} x^{n+1}}{n+1}\right]} \end{array}

\begin{array}{l}{\Rightarrow \frac{4^{n+1}-1}{n+1}=3 C_{0}+\frac{3^{2} C_{1}}{2}+\frac{3^{3} C_{2}}{3}+\frac{3^{4} C_{3}}{4}+\ldots+\frac{3^{n+1} C_{n}}{n+1}} \\\\ {\text { Hence, }} \\\\ {3 C_{0}+\frac{3^{2} C_{1}}{2}+\frac{3^{3} C_{2}}{3}+\frac{3^{4} C_{3}}{4}+\ldots+\frac{3^{n+1} C_{n}}{n+1}=\frac{4^{n+1}-1}{n+1}}\end{array}

Option C is correct.

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Divya Prakash Singh

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