expansâo binomial

sendo \( \displaystyle a_k = \binom { 2n+1}k  2^{2n+1-k} \) então

\(\displaystyle  \left( \sqrt x + 2\right)^{2n+1}  =  a_0x^0 + a_1x^{1/2} + a_2x^1 + \cdots + a_{2n+1}x^{(2n+1)/2} \)

Logo, nosso objetivo é calcular a₀ + a₂ + a₄ + ... + a_2n. Reparamos que

\(\displaystyle  \left( \sqrt x - 2\right)^{2n+1}  =  - a_0x^0 + a_1x^{1/2} - a_2x^1 + a_3x^{3/2}  - \cdots -a_{2n}x^n +  a_{2n+1}x^{(2n+1)/2} \)

Portanto:

\( \displaystyle \left( \sqrt x + 2\right)^{2n+1} -   \left( \sqrt x - 2\right)^{2n+1} = 2 (a_0x^0 + a_2x^1 + \cdots + a_{2n}x^n)\)

Assim, para x = 1 temos:

\( \displaystyle \left( \sqrt 1 + 2\right)^{2n+1} -   \left( \sqrt 1 - 2\right)^{2n+1} = 2 (a_0 + a_2 + \cdots + a_{2n}) \implies \)

\(  3^{2n+1} - (-1)^{2n+1} = 3^{2n+1} + 1  =2 (a_0 + a_2 + \cdots + a_{2n})  \implies \)

\(\boxed{ a_0 + a_2 + \cdots + a_{2n}  = \dfrac{3^{2n+1} + 1}2 }\)

nics

idem!