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If (1 + x)^(n) = C(0) + C(1) x + C(2) x...

If ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + … + C_(n) x^(n) ` , prove that
` C_(0) - (C_(1))/(2) + (C_(2))/(3) -…+ (-1)^(n) (C_(n))/(n+1) = (1)/(n+1)`.

A

0

B

`(1)/(n+ 1)`

C

`(2^(2))/(n+1)`

D

`(-1)/(n+1)`

Text Solution

Verified by Experts

The correct Answer is:
b
We have ,
`C_(0) - (C_(1))/(2) + (C_(2))/(3) - (C_(3))/(4) +...+ (-1)^(n) (C_(n))/(n+1)`
`sum_(r=0)^(n) (-1)^(r) (C_(n))/(n+1)`
`sum_(r=0)^(n) ((-1)^(r))/(r +1) .""^(n)C_(r)`
`sum_(r=0)^(n) ((-1)^(r))/(r +1).(n+1)/(r+1) .""^(n)C_(r)`
`= (1)/(n+1) sum_(r=0)^(n) (-1)^(r) .""^(n+1)C_(r+1)" "[because ""^(n+1)C_(r+1)=(n+1)/(r+1).""^(n)C_(r)]`
` (1)/(n+1) sum_(r=0)^(n) ((-1)^(r))/(r +1) .""^(n)C_(r)`
`(1)/(n+1)[""^(n+1)C_(1)-""^(n+1)C_(2)+""^(n+1)C_(3)-""^(n+1)C_(4)+...+(-1)^(n) ""^(n+1)C_(n +1)]`
`= - (1)/(n+1) [-""^(n+1)C_(1)+""^(n+1)C_(2)-""^(n+1)C_(3)+""^(n+1)C_(4)-...+(-1)^(n) ""^(n+1)C_(n +1)]`
`-(1)/(n+1) [""^(n+1)C_(0)-0""^(n+1)C_(1)+""^(n+1)C_(2)-""^(n+1)C_(3)+...+(-1)^(n+1) ""^(n+1)C_(n +1)}-""^(n+1)C_(0)]`
`= (1)/(n+1) {0 - ""^(n+1)C_(0)} = (1)/(n+1)` .
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OBJECTIVE RD SHARMA ENGLISH-BINOMIAL THEOREM AND ITS APPLCIATIONS -Chapter Test
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