1.

Show that C_0 n^2 + C_1 (2-n)^2 + C_2 (4-n)^2 + .... + C_n (2n-n)^2 = n.2^n

Answer»

SOLUTION :`C_0 n^2 + C_1 (2-n)^2 + C_2 (4-n)^2 + .... + C_n (2N-n)^2`
`n^2(C_0 + C_1 + .... + C_n) + (2^2C_1 + 4^2C_2 + ...... + (2n)^2C_n) - 4N(C_1 + 2C_2 + 3C_3 + ... + nC_n)`
= `n^2.2^n + 4n(n+1)2^(n-2) - 4n.n.2^(n-1`)
`2^(n-1)(2n^2 + 2n^2 + 2n - 4n^2)
= `2n.2^(n-1)` = `n2^n`


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