InterviewSolution
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InterviewSolution
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The value of `sum_(r=0)^(20)(-1)^(r )(""^(50)C_(r))/(r+2)` is equal toA. `(1)/(50xx51)`B. `(1)/(52xx50)`C. `1/(52xx51)`D. none of these |
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Answer» Correct Answer - C Here, `T_(r)=(-1)^(r)(.^(50)C_(r))/(r+2)` `=(-1)^(r)(r+1)(.^(50)C_(r))/((r+1)(r+2))` `=(-1)^(r)(r+1)(.^(52)C_(r+2))/(51xx52)` `= (-1)^(r)([(r+2)-1]^(52)C_(r+2))/(51xx52)` `= (-1)^(r)([52.^(51)C_(r+1)-.^(52)C_(r+2)])/(51xx52)` `= ([-52.^(51)C_(r+1)(-1)^(r+1)-.^(52)C_(r+2)(-1)^(r+2)])/(51xx52)` `underset(r=0)overset(50)sum(-1)^(r)(.^(50)C_(r))/(r+2)` ` = -52((1-1)^(51)-.^(51)C_(0))/(51xx52)-((1-1)^(52)-.^(52)C_(0)+.^(52)C_(1))/(51xx52)` `= 1/51-1/52` `= 1/(51xx52)` Alternate solution : `(1-x)^(n)=underset(r=0)overset(n)sum.^(n)C_(r)(-1)^(r)x^(r)` or `x(1-x)^(n)=underset(r=0)overset(n)sum(-1)^(r).^(n)C_(r)x^(r+1)` Intergrating both sides withing the limits 0 to 1, we get `underset(0)overset(1)intx(1-x)^(n)dx=underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2)` or `underset(r=0)overset(n)sum(-1)^(r)(.^(n)C_(r))/(r+2) = underset(0)overset(1)intx(1-x)^(n)dx` `= underset(0)overset(1)int(1-x)x^(n)dx`(Replace x by `1-x`) `= |(x^(n+1))/(n+1)-(x^(n+2))/(n+2)|_(0)^(1)` `= (1)/(n+1)-(1)/(n+2)` `= (1)/((n+1)(n+2))` Now put `n = 50`. |
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