Question

Escreva uma lei fechada para a seguinte soma

        S(n) = 1²2¹ + 2²2² + 3²2³ + ... + n²2ⁿ,        n ∈ ℕ*

e expresse a sua resposta em termos de n.

Answer 1

Olá Lukyo.

Queremos achar a fórmula fechada da soma.

$\mathsf{S_n=\displaystyle\sum_{k=1}^n 2^k\cdot k^2}$

Multiplique $\mathsf{S_n}$ por 2

$\mathsf{2S_n=\displaystyle\sum_{k=1}^n2^{k+1}\cdot k^2}$

Faça agora $\mathsf{S_n-2S_n}$

$\mathsf{S_n-2S_n=\displaystyle\sum_{k=1}^n 2^k\cdot k^2-\sum_{k=1}^{n}2^{k+1}\cdot k^2}$


Some 1 no limitante inferior e superior do segundo somatório.


$\mathsf{-S_n=\displaystyle\sum_{k=1}^n 2^k\cdot k^2-\sum_{k=2}^{n+1}2^k \cdot(k - 1)^2}$


Tire o primeiro termo do primeiro somatório para fora e tire o último termo do segundo somatório para fora.


$\mathsf{-S_n=2^1\cdot1^2+\displaystyle\sum_{k=2}^n [2^k\cdot k^2]-2^{n+1}\cdot n^2-\sum_{k=2}^n 2^k\cdot(k-1)^2}\\\\\\\\\mathsf{-S_n=2-2^{n+1}\cdot n^2+\displastyle\sum_{k=2}^n 2^k\cdot [k^2-(k-1)^2]}\\\\\\\\\mathsf{-S_n=2-2^{n+1}\cdot n^2+\displaystyle\sum_{k=2}^n 2^k\cdot (2k-1)}\\\\\\\\\mathsf{-S_n=2-2^{n+1}\cdot n^2+\displaystyle\sum_{k=2}^n 2^{k+1}\cdot k-\sum_{k=2}^n 2^k}$


Vamos achar agora fórmula fechada da seguinte série.


$\mathsf{T_n=\displaystyle\sum_{k=2}^n 2^{k+1}\cdot k}$


Multiplique $\mathsf{T_n}$ por 2.


$\mathsf{2T_n=\displaystyle\sum_{k=2}^n 2^{k+2}\cdot k}$


Faça $\mathsf{2T_n-T_n}$


$\mathsf{2T_n-T_n=\displaystyle\sum_{k=2}^n 2^{k+2}\cdot k -\sum_{k=2}^n 2^{k+1}\cdot k}\\\\\\\\\mathsf{T_n=\displaystyle\sum_{k=2+1}^{n+1}2^{k+1}\cdot (k-1)-\sum_{k=2}^n 2^{k+1}\cdot k}\\\\\\\\\mathsf{T_n=2^{n+2}\cdot n+\displaystyle\sum_{k=3}^n [2^{k+1}\cdot (k - 1)]-2^3\cdot 2-\sum_{k=3}^n 2^{k+1}\cdot k}\\\\\\\\\mathsf{T_n=2^{n+2}\cdot n-16-\displaystyle\sum_{k=3}^{n}2^{k+1}}$


Note que apareceu a soma de uma P.G. de razão 2.


$\mathsf{\displaystyle\sum_{k=3}^n 2^{k+1}=2^4 \cdot\dfrac{(2^{n-2}-1)}{2-1}=(2^{n+2}-16)}$


Substituindo em $\mathsf{T_n}$


$\mathsf{T_n=2^{n+2}\cdot n-16-(2^{n+2}-16)}\\\\\\\mathsf{T_n=2^{n+2}\cdot n - 16-2^{n+2}+16}\\\\\\\mathsf{T_n=\displaystyle\sum_{k=2}^n 2^{k+1}\cdot k =2^{n+2}\cdot(n-1)}$


Portanto $\mathsf{S_n}$ fica:


$\mathsf{-S_n=2-2^{n+1}\cdot n^2+\displaystyle\sum_{k=2}^n 2^{k+1}\cdot k-\sum_{k=2}^n 2^k}\\\\\\\\\mathsf{-S_n=2-2^{n+1}\cdot n^2+[2^{n+2}\cdot (n-1)]-\Big[2^2\cdot\dfrac{(2^{n-1}-1)}{2-1}\Big]}\\\\\\\\\mathsf{-S_n=2-2^{n+1}\cdot n^2+2^{n+2}\cdot n-2^{n+2}-2^{n+1}+4}\\\\\\\\\boxed{\mathsf{S_n=2\cdot(2^{n}\cdot n^2-2^{n+1}\cdot n-3\cdot2^{n+1}+3)}}$


Dúvidas? Comente.