InterviewSolution
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InterviewSolution
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Prove that `.^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2+1/3) .^(n)C_(3) + "…" + (-1)^(n-1) (1+1/2+1/3 + "…." + 1/n) .^(n)C_(n) = 1/n` |
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Answer» Let `S = .^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2 + 1/3) .^(n)C_(3) + "….."` `+(-1)^(n-1)(1+1/2+1/3+"…."+1/n) .^(n)C_(n)`. rth term of the series is `T(r ) = (-1)^(r-1) . C_(r ) (1+1/2+1/3+"……"+1/r)`, where `C_(r) = .^(n)C_(r)` Let us consider a series whose general term is `T_(1)(r ) = (-1)^(r-1). C_(r)(1+x+x^(2)+"...."+x^(r-1))` `= (-1)^(r-1). C_(r) ((1-x^(r))/(1-x))` ` = ((-1)^(r-1)C_(r))/(1-x) + ((-1)^(r)x^(r)C_(r))/(1-x)` `rArr underset(r=1)overset(n)sum T_(1)(r ) = (1)/((x-1)) underset(r=1)overset(n)sum (-1)^(r) C_(r ) + (1)/((1+x))underset(r=1)overset(n)sum(-1)^(r). C_(r).x^(r )` `rArr underset(r=1)overset(n)sumT_(1)(r ) = (1)/((x-1)) (0-x) + (1)/((1-x)) ((1-x)^(n) - 1) = (1-x)^(n-1)` `= L` (say) Clearly `S = underset(0 )overset(1)intLdx = underset(0)overset(1)int(1-x)^(n-1) dx = 1/n` |
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