InterviewSolution
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InterviewSolution
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Prove that `(.^(2n)C_(0))^(2) - (.^(2n)C_(1))^(2) + (.^(2n)C_(2))^(2) - …. + (.^(2n)C_(2n))^(2) = (-1)^(n) .^(2n)C_(n)` |
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Answer» `(1+x)^2n(1-(1)/(x))^2n` `=(.^2nC_0 -(.^2n C_1)x+(.^2nC_2)x62+.....+(.^2nC_2n)x^2n]` `xx[.^2nC_0 -(.^2nC_1)(1)/(X)+(.^2n C_2)(1)/(x^2)+.....+(.^2n C_2n)(1)/(x^2n)]` Independent terms of x on RHS `=(.^2n C_0)^2-(.^2n C_1)^2+(.^2nC_2)^2 -......+(.^2n C_2n)^2`. LHS `=(1+x)^2n((x-1)/(x))^2n =(1)/(x^2n)(1-x^2)^2n`. Independent term of x on the LHS `=(-1)^n .^2n C_n`. |
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