2s=2^2+2^3+2^4+2^5+....+2^201
lấy 2s-s ta có:(2^2+2^3+2^4+2^5+....+2^201)-(2+2^2+2^3+2^4+.....+2^200)
\(\Rightarrow\)s=2^201-2
3q=3^2-3^3+3^4-.....-3^2017+3^2018
3q+q=3+3^2018
4q=3+3^2018
q=\(\dfrac{3+3^{2018}}{4}\)
\(=>2S=2^2+2^3+2^4+......+2^{200}+2^{201}\)
\(=>2S-S=2^{201}-2\)
\(=>S=2^{201}-2\)
b , Ta có:
\(Q=3\left(1-3\right)+3^3\left(1-3\right)+.....+3^{2015}\left(1-3\right)+3^{2017}\)
\(=>Q=-2\left(3+3^3+3^5+......+3^{2015}\right)+3^{2017}\)
\(=>9Q=-2\left(3^3+3^5+......+3^{2015}+3^{2017}\right)+3^{2019}\)
\(=>9Q-Q=-2.3^{2017}+3^{2019}-2.3-3^{2017}\)
\(=>8Q=3^{2017}\left(-2+3^2-1\right)-6\)
\(=>8Q=3^{2017}.6-6\)
\(=>Q=\dfrac{3^{2017}.6-6}{8}=\dfrac{3^{2017}.3-3}{4}=\dfrac{3^{2018}-3}{4}\)
CHÚC BẠN HỌC TỐT..........
