Giải thích các bước giải:
Ta có:
$u_n=-\dfrac32\cdot 3^{n-1}+\dfrac92-n=-\dfrac12\cdot 3^n+\dfrac92-n$
Ta có:
$S=u_1+u_2+...+u_{20}$
$\to S=(-\dfrac12\cdot 3^1+\dfrac92-1)+(-\dfrac12\cdot 3^2+\dfrac92-2)+...+(-\dfrac12\cdot 3^{20}+\dfrac92-20)$
$\to S=-\dfrac12(3^1+3^2+...+3^{20})+20\cdot(\dfrac92-1)$
$\to S=-\dfrac12\cdot \dfrac{3^{21}-3}{3-1}+20\cdot\dfrac72$
$\to S=-\dfrac12\cdot \dfrac{3^{21}-3}{2}+70$
$\to S=- \dfrac{3^{21}-3}{4}+70$