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)\(P=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3P=3\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}\right)\)
\(3P=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(3P-P=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(2P=1-\frac{1}{3^{100}}\)
\(P=\left(1-\frac{1}{3^{100}}\right):2\)
S=2+22+23+...+2100
=>2S=22+23+24+...+2101
=>2S-S=(22+23+24+...+2101)-(2+22+23+...+2100)
=>S=2101-2=2(2100-1)
\(F=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3F=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3F-F=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(\Rightarrow2F=1-\frac{1}{3^{100}}\)
\(\Rightarrow F=\frac{1-\frac{1}{3^{100}}}{2}\)
Câu P đó, bạn nhân 3 rồi tính 3P-P, xong rút gọn rồi ra thôi
Tính
\(A=\frac{0,4.10^{-8}.4.6,28}{3,14.\left(2.10^{-5}\right)^2}\)