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If C_(r) = .^(n)C_(r) then prove that (C_(0) + C_(1)) (C_(1) + C_(2)) "….." (C_(n-1) + C_(n)) = (C_(1)C_(2)"…."C_(n-1)C_(n))(n+1)^(n)//n! |
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Answer» Solution :We have, `(C_(0)+C_(1))(C_(1)+C_(2))(C_(2)+C_(3))"...."(C_(N-1)+C_(n))` `=C_(1)C_(2)"...."C_(n-1)C_(n)(1+(C_(0))/(C_(1)))(1+(C_(1))/(C_(2)))(1+(C_(2))/(C_(3)))"....."(1+(C_(n-1))/(C_(n)))` `=C_(1)C_(2)"...."C_(n-1)C_(n)(1+1/n)(1+2/(n-1))(1+3/(n-2))"...."(1+n/1)` `=C_(1)C_(2)"....."C_(n-1)C_(n)((n+1)^(n))/(n!)` |
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