Đáp án:
lim$\dfrac{1+2+2^{2}+...+2^{n}}{1+3+3^{2}+...+3^{n}}$=0
Giải thích các bước giải:
Ta có:
Đặt A=$1+2+2^{2}+...+2^n$
2A=$2+2^2+2^{3}+...+2^{n+1}$
⇒A=2A-A=$2^{n+1}-1$
Đặt B=$1+3+3^2+...+3^{n}$
3B=$3+3^2+3^{3}+...+3^{n+1}$
⇒3B-B=2B= $3^{n+1}-1$
⇒B= $\dfrac{3^{n+1}-1}{2}$
⇒lim$\dfrac{1+2+2^{2}+...+2^{n}}{1+3+3^{2}+...+3^{n}}$
=lim$\dfrac{2^{n+1}-1}{\dfrac{3^{n+1}-1}{2}}$
= lim$\dfrac{2.(2^{n+1}-1)}{3^{n+1}-1}$
=lim$\dfrac{2.(2^{n}.2-1)}{3^{n}.3-1}$
=lim$\dfrac{4.2^{n}-2}{3.3^n-1}$
=lim$\dfrac{2^n.(4-\dfrac{2}{2^n})}{3^n.(3-\dfrac{1}{3^n})}$
=lim$\dfrac{2^n}{3^n}$
=lim$(\dfrac{2}{3})^{n}$
=0
