\(N=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(N=\dfrac{1}{2^1}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\)
\(2N=1-\dfrac{1}{2^1}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\)
\(2N+N=1-\dfrac{1}{2^6}\)
\(N=\dfrac{1}{3}-\dfrac{1}{2^6.3}< \dfrac{1}{3}\left(đpcm\right)\)
Tính tổng sau đây bằng 2 cách:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
CMR : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Với \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)
Bài 1: tính
Cho A= \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+........+\dfrac{1}{60}>\dfrac{7}{12}\)
B=\(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{50^2}\)
CMR B > \(\dfrac{1}{4}\); B < \(\dfrac{4}{9}\)
C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}...........\dfrac{79}{80}\)<\(\dfrac{1}{9}\)
Chứng mình rằng :
a) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \)\(\dfrac{1}{2}\)
b)\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)
Giup mk nha ! Đang cần gấp lắm rùi !
Tìm x :
\(\dfrac{1}{4}\) . \(\dfrac{2}{6}\) . \(\dfrac{3}{8}\) . \(\dfrac{4}{10}\) . .... . \(\dfrac{30}{62}\) . \(\dfrac{31}{64}\)
Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\)< \(\dfrac{3}{16}\)
Tính tổng sau:
a) \(\dfrac{1}{9}+3,25+5\dfrac{3}{16}+4\dfrac{1}{3}+2,8+0,5\)
b) \(2\dfrac{1}{3}+0,45+4,25+\dfrac{1}{81}+6\dfrac{8}{27}\)
c) \(1,25+2\dfrac{1}{4}+4\dfrac{2}{5}+0,3+2,14+4\dfrac{1}{8}\)
CMR \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
