Question

Tính :

\(S=2+2^2+2^3+...+2^{100}\)

\(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

Answer 1

a)S=2+2²+2³+...+2¹⁰⁰

2S=2(2+2²+2³+...+2¹⁰⁰)

2S=2²+2³+...+2¹⁰¹

2S-S=(2²+2³+...+2¹⁰¹)-(2+2²+2³+...+2¹⁰⁰)

S=2¹⁰¹-2

b)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)

\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)

\(2P=1-\frac{1}{3^{100}}\)

\(P=\left(1-\frac{1}{3^{100}}\right):2\)