1.

Prove that "^(2n)C_0 + ^(2n)C_2 + .... + ^(2n)C_(2n) = 2^(2n-1)

Answer»

Solution :`"^(2N)C_0 + ^(2n)C_2 + .... + ^(2n)C_(2n)` = 2^(2n-1)`
We know that `(1+X)^(2n) = "^(2n)C_0 + ^(2n)C_1 x + ^(2n)C_2 x^2 + ... ^(2n)C_(2n) x^n` ... (1)
Putting x = 1
we get putting x = 1 we get
`("^(2n)C_0 + ^(2n)C_2 + ^(2n)C_4 + .... + ^(2n)C_(2n)) - (^(2n)C_1 + ^(2n)C_3 + ... + ^(2n)C_(2n-1))` = 0
therefore ^(2n)C_0 + ^(2n)C_2 + .... + ^(2n)C_(2n)` = `"^(2n)C_1 + ^(2n)C_3 + .... + ^(2n)C_(2n-1)`
= `2^(2n)/2` = `2^(2n-1)`


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