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Prove that .^(n)C(0) + (.^(n)C(1))/(2) ...

Prove that `.^(n)C_(0) + (.^(n)C_(1))/(2) + (.^(n)C_(2))/(3) + "……" +(. ^(n)C_(n))/(n+1) = (2^(n+1)-1)/(n+1)`.

Text Solution

Verified by Experts

Method I :
We have
`T_(r ) = (.^(n)C_(r-1))/(r ) = (.^(n+1)C_(r))/(n+1)`
`:.` Required sum `= underset(r=1)overset(n+1)sumT_(r)`
` = underset(r=1)overset(n+1)sum(.^(n+1)Cr)/(n+1)`
` = (.^(n+1)C_(1) + .^(n+1)C_(2) + "….." + .^(n+1)C_(n+1))/(n+1)`
` = ((.^(n+1)C_(0) + .^(n+1)C_(1) + .^(n+1)C_(2) + "......" + .^(n+1)C_(n+1))-1)/(n+1)`
`= (2^(n+1) - 1)/(n+1)`
Method II :
We have `(1+x)^(n)= .^(n)C_(0) + .^(n)C_(1)x+.^(n)C_(2)x^(2)+"...."+.^(n)C_(n)x^(n)`
` :. underset(0)overset(1)int(1+x)^(n)dx = underset(0)overset(1)int(.^(n)C_(0) + .^(n)C_(1)x+.^(n)C_(2)x^(2) + "...." + .^(n)C_(n)x^(n))dx`
`rArr [((1+x)^(n+1))/(n+1)]_(0)^(1)=[.^(n)C_(0)x+(.^(n)C_(1)x^(2))/(2)+(.^(n)C_(2)x^(3))/(3) + "....."+ (.^(n)C_(n)x^(n+1))/(n+1)]_(0)^(1)`
`rArr (2^(n+1))/(n+1) -1/(n+1)=.^(n)C_(0)+(.^(n)C_(1))/(2)+(.^(n)C_(2))/(3) + "......" + (.^(n)C_(n))/(n+1)`
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