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The sum to (n + 1) terms of the 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

As we have learned

Result of Binomial Theorem -

c_{0}+\frac{c_{1}}{2}+\frac{c_{2}}{3}+-----\frac{c_{n}}{n+1}= \frac{2^{n+1}-1}{n+1}

 

- wherein

Take \int_{0}^{1}\left ( 1+x \right )^{n}dx= \int_{o}^{1}\sum_{r=0}^{n}\, ^{n}c_{r}\, x^{r}dx

 

 

d

                              We have

                                    (1 -x)n = C0-C1x + C2x2-C3x3 + …

                              Þ   x(1 -x)n = C0x-C1x2 + C2x3-C3x4 + …

                                    \int_{0}^{1} x (1-x)^n dx = \int_{0}^{1} x (1-x )^ndx = \int_{0}^{1} (1-t) t^n (-1)dt \: \: \: \: [ put \: \: 1-x = t ]     =    [Put 1 -x = t]

                                                            = \int_{0}^{1} ( t^n -t^{n+1})dt  = \left | \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2}\right |_{0}^{1}

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

                              Integrating R.H.S of (1) we get \left ( \frac{C_0 x ^2 }{}2 - \frac{C_1x^3 }{}3+ \frac{C_2 x^4}{4}- \right )|_{0}^{1}

                                                            = \frac{C_0 }{}2 - \frac{C_1 }{}3+ \frac{C_2 }{4}-.... 

Thus                     

                                    \frac{C_0 }{}2 - \frac{C_1 }{}3+ \frac{C_2 }{4}-....           = \frac{1}{(n+1)(n+2)}

 

 

 

 

 

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