How find this sum_(k=0)^(2n)(−1)^k a^2_k

Jazmyn Pugh

Jazmyn Pugh

Answered question

2022-09-22

How find this k = 0 2 n ( 1 ) k a k 2

Answer & Explanation

Klecanlh

Klecanlh

Beginner2022-09-23Added 11 answers

The form of the polynomial p ( x ) = 1 2 x + x 2 appeals to use its conjugate polynomial q ( x ) = 1 2 x + x 2 . Such the usage can be a evaluation of roots and applying Vieta’s formulas, or dealing with a recurrent sequence. But to chose the relevant way to solve the problem we should begin from the end (as recommended George Polya). So, we can see that the asked sum S = k = 0 2 n ( 1 ) k a k 2 is equal to the coefficient at x 0 of the “polynomial” p n ( x ) p n ( 1 / x ). It rests to find it. We have
p n ( x ) p n ( 1 / x ) = ( p ( x ) p ( 1 / x ) ) n =
( ( 1 2 x + x 2 ) ( 1 / x 2 + 2 / x + 1 ) ) n =
( ( 1 / x 2 ) ( 1 2 x + x 2 ) ( 1 + 2 x + x 2 ) ) n =
( ( 1 / x 2 ) ( ( 1 + x 2 ) 2 ( 2 x ) 2 ) ) n = ( x 2 + 1 / x 2 ) n .
So we see that S=0provided n is odd and S = ( n n / 2 ) provided n is even.

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