InterviewSolution
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InterviewSolution
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If `sum_(r=0)^(2n)(-1)^r(2n(C_r)^2=alpha_1`, then `sum_(r=0)^(2n)(-1)^r(r-2n)(2n(C_r)^2=`A. `nalpha`B. `-n alpha `C. `0D. none of these |
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Answer» Correct Answer - b Let `beta = sum_(r=0)^(2n) (-1)^(r) (r -2n)(""^(2n)C_(r))^(2)` …(i) `rArr beta = - sum_(r=0)^(2n) (-1)^(r) (2n -r) (""^(2n)C_(2n-r))^(2)` Writing the terms in the neverse order, we get `beta = sum_(r=0)^(2n) (-1)^(r) r(""^(2n)C_(r))^(2) ` ...(ii) Adding (i) and (ii), we get `2beta = - sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)` `rArr 2 beta = - 2n sum_(r=0)^(2n) (-1)^(r) (""^(2n)C_(r))^(2)` `rArr 2 beta = - 2nalpha rArr beta = - n alpha ` . |
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