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Prove that 1/(n+1)=(.^n C1)/2-(2(.^n C2)...

Prove that `1/(n+1)=(.^n C_1)/2-(2(.^n C_2))/3+(3(.^n C_3))/4- . . . +(-1^(n+1))(n*(.^n C_n))/(n+1)` .

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`S = (.^(n)C_(1))/(2)-(2(.^(n)C_(3)))/(3)+(3(.^(n)C_(3)))/(4)"...."+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)`
` = underset(r=1)overset(n)sum (r.^(n)C_(r))/((r+1))(-1)^(r+1)`
`= underset(r=1)overset(n)sum r(.^(n+1)C(r+1))/(n+1)(-1)^(r+1)`
`= 1/(n+1) underset(r=1)overset(n)sum [(r+1).^(n+1)C_(r+1)(-1)^(r+1)-.^(n+1)C_(r+1)(-1)^(r+1)]`
`= (1)/(n+1)underset(r=1)overset(n)sum[-(n+1).^(n)C_(r)(-1)^(r)-.^(n+1)C_(r+1)(-1)^(r+1)]`
`= -[-.^(n)C_(2)-.^(n)C_(3)+"...."+(-1)^(n).^(n)C_(n)]`
` - 1/(n+1)[.^(n+1)C_(2) - .^(n+1)C_(3)+".....(-1)^(n+1).^(n+1)C_(n+1)]`
`= - [(1-1)^(n)-1]-(1)/(n+1)[(1-1)^(n+1) - .^(n+1)C_(0)+.^(n+1)C_(1)]`
` = 1 - (1)/(n+1)[(n+1)-1]`
`1 - (n)/(n+1)= (1)/(n+1)`
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