\(CMR:\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+................+\frac{99}{100!}<1\)
Những câu hỏi liên quan
CMR: \(\frac{1}{3}-\frac{2}{3^2}=\frac{3}{3^3}-\frac{4}{4^4}+.......+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....\frac{1}{3^{98}}-\frac{1}{3^{99}}\Rightarrow4A< B\left(1\right)\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+....\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\Rightarrow4B< 3\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4A< B< \frac{3}{4}\Rightarrow4A< \frac{3}{4}\Rightarrow A< \frac{3}{4}:4\Rightarrow A< \frac{3}{4}.\frac{1}{4}\Rightarrow A< \frac{3}{16}\)
=> đpcm.
CMR \(\frac{1}{3}-\frac{2}{3^2}+\frac{2}{3^3}-\frac{4}{3^4}+......+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}
Đặt A=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) +\(\frac{3}{3^3}\) - \(\frac{4}{3^4}\)+...+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3A=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\)+...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4A=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)- \(\frac{100}{3^{100}}\)
=> 4A<1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\) (1)
Đặt B=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)
=> B=2+ \(\frac{1}{3}\) - \(\frac{1}{3^2}\) +...+\(\frac{1}{3^{97}}\) - \(\frac{1}{3^{98}}\)
=> 4B=B+3B=3-\(\frac{1}{3^{99}}\)<3 => A<\(\frac{3}{4}\) (2)
Từ (1) và (2) ta có: 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
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CMR\(\frac{1}{3^1}-\frac{1}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{16}\)
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(3A+A=4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4A< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(B+3B=4B=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow B< \frac{3}{4}\) (2)
Từ (2) và (2) => \(4A< B< \frac{3}{4}\Rightarrow A< \frac{3}{16}\) (đpcm)
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\(A=\frac{7n-1}{4};B=\frac{5n+3}{12}\)
Tìm n để A,B đồng thời là các số nguyên tố
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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}
Đọc tiếp
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. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
CMR:\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{16}\)\(\frac{3}{16}\)
Nhầm đầu bài nhoa:
Phải là \(-\frac{100}{3^{100}}\)
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2.CMR:
\(S2=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
CMR S=\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}^{ }<\frac{3}{16}\)
Lớp 6
CMR : \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)