InterviewSolution
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InterviewSolution
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Statement -2: `sum_(r=0)^(n) (-1)^( r) (""^(n)C_(r))/(r+1) = (1)/(n+1)` Statement-2: ` sum_(r=0)^(n) (-1)^(r) (""^(n)C_(r))/(r+1) x^(r) = (1)/((n+1)x) { 1 - (1 - x)^(n+1)}`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - a We have, `sum_(r=0)^(n) (-1)^(r) (""^(n)C_(r))/(r+1)x^(r)` ` = - (1)/(x(x +1)) sum_(r=0)^(n) (-1)^(r+1) (n+1)/(r+1) ""^(n)C_(r) x^(r+1)` `= - (1)/(x (n+1)) {( 1 - x)^(n +1) -1} = (1)/((n+1)x) {1 - (1 - x)^(n +1)}` ` = - (1)/(x(n+1)) {(1 -x)^(n+1) -1} = (1)/((n+1)x) {1 - ( 1 - x)^(n+1)}` So, statement-2 is true Replacing x by 1 in statement-2, we get `sum_(r=0)^(n) (-1)^^(r) (""^(n)C_(r))/(r+1) = (1)/(n+1)` So, statement-1 is also true and stetement-2 is a correct explanation for statement-1. |
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