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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)`. |
Answer» `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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