InterviewSolution
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InterviewSolution
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Prove that `C_0 – 2^2 C_1 + 3² C_2 – 4^2 C_3 + ... +(-1)^n (n + 1)^2 C_n = 0` where `C_r = nC_r` |
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Answer» `C_(0) - 2^(2) C_(1) + 3^(2) . C_(2) - … + (-1)^(n) (n+1)^(2) . C_(n)` `= sum_(r = 0 )^(n) *(-1)^(r) (r + 1)^(2)""^(n)C_(r) = sum_(r= 0) ^(n) (-1)^(r)(r^(2) + 2r + 1 ) ""^(n)C_(r)` `= sum_(r = 0 )^(n) *(-1)^(r)r^(2). ""^(n)C_(r) +2 sum_(r= 0) ^(n) (-1)^(r)r. ""^(n)C_(r) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)` `= sum_(r = 0 )^(n) *(-1)^(r)r(r - 1) ""^(n)C_(r) +3 sum_(r= 0) ^(n) (-1)^(r)r. ""^(n)C_(r) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)` `= sum_(r = 0 )^(n) *(-1)^(r)n(n - 1) ""^(n - 2)C_(r-2) +3 sum_(r= 0) ^(n) (-1)^(r)n. ""^(n-1)C_(r-1) + sum_(r=0)^(n) (-1)^(r). ""^(n)C_(r)` ` = n(n- 1) {""^(n-2)C_(0) - ""^(n - 2)C_(0) - ""^(n-2)C_(1) + ""^(n-2)C_(2^(-))...+(-1)^(n) ""^(n-2)C_(n-2)}` ` = + 3n{-""^(n-1)C_(0)+ ""^(n - 1)C_(1) - ""^(n-1)C_(2)+ ...+(-1)^(n) ""^(n-1)C_(n-1)} + {""^(n)C_(0) -""^(n)C_(1) + ""^(n)C_(2) +...+ (-1)^(n)""^(n)C_(n)} `= n(n-1). 0 + 3n .0 - AA n gt 2 = 0 , AA n gt 2 ` . |
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