If `k a n d n` are positive integers and `s_k=1^k+2^k+3^k++n^k ,` then prove that `sum_(r=1)^m^(m+1)C_r s_r=(n+1)^(m+1)-(n+1)dot`

Explanation

`S = underset(r=1)overset(m)sum.^(m+1)C_(r) S_(r)`
`= [.^(m+1)C_(1)s_(1)+.^(m+1)C_(2)s_(2) + "……." + .^(m+1)C_(m)s_(m)]`
`= .^(m+1)C_(1)(1+2+3+"….."+n)`
`+ .^(m+1)C_(2)(1^(3)+2^(3)+3^(3)+"……."+n^(3))`
`+.^(m+1)C_(3)(1^(3)+2^(3)+3^(3)+"......."+n^(m))`
`+.^(m+1)C_(m)(1^(m)+2^(m)+3^(m)+"....."+n^(m))`
`= (.^(m+1)C_(1)1+.^(m+1)C_(2)1^(2)+.^(m+1)C_(3)1^(3) + "......."+.^(m+1)C_(m)1^(m))`
`+(.^(m+1)C_(1)2+.^(m+1)C_(2)2^(2)+.^(m+1)C_(3)2^(3)+"......." +.^(m+1)C_(m)2^(m))+"......."+(.^(m+1)C_(1)n+.^(m+1)C_(2)n^(2)+"......"+.^(m+1)C_(m)n^(m))`
` = [(1+1)^(m+1)-1-.^(m+1)C_(m+1)1^(m+1)]`
`+[(1+2)^(m+1)-1-.^(m+1)C_(m+1)2^(m+1)]`
`+[(1+3)^(m+1)-1-.^(m+1)C_(m+1)3^(m+1)]+"......"`
` = (2^(m+1)-1^(m+1))+(3^(m+1)-2^(m+1))+(4^(m+1)-3^(m+1))+"......"`
`+ [(1+n)^(m+1)-n^(m+1)]-n`
`= (1+n)^(m+1)-1-n=(1+n)^(m+1)- (n+1)`