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If (1+x)^n=C0+C1\ x+C2\ x^2+\ +Cn\ x^n ,...

If `(1+x)^n=C_0+C_1\ x+C_2\ x^2+\ +C_n\ x^n` , using derivatives prove that
(I)`C_1+2C_2+\ dot+n C_n=n .2^(n-1)`
(ii) `C_1-2C_2+3C_3+\ dot+(-1)^(n-1)\ n C_n=0`

Text Solution

Verified by Experts

Given that,
`(1+x)^n=C_0+C_1\ x+C_2\ x^2+\ +C_n\ x^n`

(i) `(1+x)^n=C_0+C_1\ x+C_2\ x^2+\ +C_n\ x^n`

Differentiating w.r.t `x` both sides

`n(1+x)^(n-1)=0+C_1+2C_2\ x+\ \...+nC_n\ x^(n-1)`

Putting `x=1`

`n(1+1)^(n-1)=0+C_1+2C_2\ (1)+\ \...+nC_n\ (1)^(n-1)`

`n(2)^(n-1)=C_1+2C_2+\ \...+nC_n\ `


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