Biết rằng S=1+2.3+3.3^2+....+11.3^10=a+21.3^b/4

* Hướng dẫn giải

Chọn C

Từ giả thiết suy ra $3S=3+{2.3}^2+{3.3}^3+\mathrm{...}+{11.3}^{11}$. Do đó 

$\begin{array}{l}-2S=S-3S\\ =1+3+3^2+\mathrm{...}+3^{10}-{11.3}^{11}\\ =1.\frac{1-3^{11}}{1-3}-{11.3}^{11}\\ =-\frac{1}{2}-\frac{{21.3}^{11}}{2}\\ \Rightarrow S=\frac{1}{4}+\frac{21}{4}{.3}^{11}.\end{array}$

vì 

$\begin{array}{l}S=\frac{1}{4}+\frac{{21.3}^{11}}{4}=a+\frac{{21.3}^b}{4}\\ \Rightarrow a=\frac{1}{4},b=11\\ \Rightarrow P=\frac{1}{4}+\frac{11}{4}=3.\end{array}$