Cho dãy số ( $U_{n}$ ) thỏa mãn $\left \{ {{U_{n}=2019} \atop {U_{n+1} = U_{n} + 2^{n} + 2020}} \right.$ ∀ n ∈ N*
Tính tổng $S_{n}$ = $U_{1}$ + $U_{2}$ + … + $U_{n}$
Cho dãy số ( $U_{n}$ ) thỏa mãn $\left \{ {{U_{n}=2019} \atop {U_{n+1} = U_{n} + 2^{n} + 2020}} \right.$ ∀ n ∈ N*
Tính tổng $S_{n}$ = $U_{1}$ + $U_{2}$ + … + $U_{n}$
Đáp án: $ S_{n}=(2^{n+1}-2)+2020\cdot \dfrac{n(n+1)}{2}-2\cdot n$
Giải thích các bước giải:
Ta có: $U_{n+1}=U_n+2^n+2020$
$\to U_{n+1}-U_n=2^{n}+2020$
$\to\begin{cases}U_n-U_{n-1}=2^{n-1}+2020\\ U_{n-1}-U_{n-2}=2^{n-2}+2020\\\cdots\\\cdots\\ U_2-U_1=2^1+2020\\ U_1=2020\end{cases}$
Cộng vế với vế
$\to U_n=(2^{n-1}+2020)+(2^{n-2}+2020)+…+(2^1+2020)+2020$
$\to U_n=(2^{n-1}+2^{n-2}+…+2^1)+2020n$
$\to U_n=(2^{n}-2)+2020n$
$\to U_n=2^{n}+2020n-2$
$\to S_n=(2^1+2020\cdot 1-2)+(2^2+2020\cdot 2-2)+…+(2^n+2020\cdot n-2)$
$\to S_{n}=(2^1+2^2+…+2^n)+2020\cdot (1+2+..+n)-2\cdot n$
$\to S_{n}=(2^{n+1}-2)+2020\cdot \dfrac{n(n+1)}{2}-2\cdot n$