Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Tính tổng :
1, A = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.................+\frac{1}{100}\)
2, B = \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+....................+\frac{99}{100}\)
Rút gọn A=\(\frac{\left(1+2+3+......+99+100\right).\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right).\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+.........+99-100}\)=...
Rút gọn B= \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+...+99-100}\)
\(1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+..+\frac{100}{2^{100}}\)
A=\(1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
Rút gọn A=\(\frac{\left(1+2+3+.......+99+100\right).\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right).\left(63.1,2-21.36+1\right)}{1-2+3-4+5-6+....+99-100}\)là .....
tính :A =1+\(\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
Tính A=1+\(\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}\)