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

Answer

N/a

Explanation

`L.H.S. =C_(0) +(C_(1))/(2)+(C_(2))/(3)+….+(C_(n))/(n+1)`
`=1+(n)/(2)+(n(n-1))/(|ul2.3)+....+(1)/(n+1)`
` =(1)/(n+1)[(n+1)+((n+2)n)/(|ul2)`
`+((n+1)n(n-1))/(|ul3)+.....+1]`
`=(1)/(n+1)[{1+(n+1)+((n+1)n)/(|ul2)`
`+((n+1)n(n-1))/(|ul3)+......+1}-]`
`=(1)/(n+1)[{.^(n+1)C_(0)+^(n+1)C_(1)+^(n+1)C_(2)`
`+^(n+1)C_(3)+......+^(n+1)C_(n+1)}-1]`
`=(1)/(n+1)[2^(n+1)-1]`
`=(2^(n+1)-1)/(n+1)=R.H.S.` Hence Proved