1. Cho S1=1+\(\frac{1}{5}\)
S2=1+\(\frac{1}{5}+\frac{1}{5^2}\)
.....
Sn=\(1+\frac{1}{5}+.....\frac{1}{5^n}\)
CMR:\(\frac{1}{5.S_1^2}+\frac{1}{5^2.S_2^2}+....+\frac{1}{5^n.S^2_n}< \frac{1}{4}\)
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Cho \(S_1=1+\frac{1}{5}\), \(S_2=1+\frac{1}{5}+\frac{1}{5^2}\), \(S_3=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\), tới \(S_n=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...........+\frac{1}{5^n}\). Chứng minh rằng : \(A=\frac{1}{5S_1^2}+\frac{1}{5^2S_2^2}+\frac{1}{5^3S_3^2}+\frac{1}{5^4S_4^2}+..........+\frac{1}{5^nS_n^2}<\frac{35}{36}\)
Khi \(n=1\to A=\frac{1}{5S_1^2}=\frac{5}{36}S_{k-1}\to S^2_k>S_k\cdot S_{k-1}\).
Vậy ta có \(\frac{1}{5^kS_k^2}
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Cho S_1-S_2+S_3-S_4+S_5frac{m}{n} với m, n nguyên tố cùng nhau. Biết:S_1frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}S_2frac{1}{2cdot3}+frac{1}{2cdot4}+frac{1}{2cdot5}+frac{1}{2cdot6}+frac{1}{3cdot4}+frac{1}{3cdot5}+frac{1}{3cdot6}+frac{1}{4cdot5}+frac{1}{4cdot6}+frac{1}{5cdot6}S_3frac{1}{2cdot3cdot4}+frac{1}{2cdot3cdot5}+frac{1}{2cdot3cdot6}+frac{1}{2cdot4cdot5}+frac{1}{2cdot4cdot6}+frac{1}{2cdot5cdot6}+frac{1}{3cdot4cdot5}+frac{1}{3cdot4cdot6}+frac{1}{3cdot5cdot6}+frac{1}{4cdot5cdot6}...
Đọc tiếp
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
CMR: \(\frac{1}{5}< \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{7}< \frac{2}{5}\)
CMR
\(\frac{1}{5}< \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)\(\frac{2}{5}\)
CMR:\(\frac{1}{5}< \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)
Cho S1=1+\(\frac{1}{5}\);S2=1+\(\frac{1}{5}+\frac{1}{5^2};...;\)Sn=\(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^n}\left(n\inℕ^∗\right)\)
CMR:1/5S12+1/52S22+.
..+1/5nSn2<1/4
Cho S1=1+\(\frac{1}{5}\);S2=1+\(\frac{1}{5}+\frac{1}{5^2};...;\)Sn=\(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^n}\left(n\inℕ^∗\right)\)
CMR:1/5S12+1/52S22+.
..+1/5nSn2<1/4
Cho S1=1+\(\frac{1}{5}\);S2=1+\(\frac{1}{5}+\frac{1}{5^2};...;\)Sn=\(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^n}\left(n\inℕ^∗\right)\)
CMR:1/5S12+1/52S22+.
..+1/5nSn2<1/4
Cho S1=1+\(\frac{1}{5}\);S2=1+\(\frac{1}{5}+\frac{1}{5^2};...;\)Sn=\(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^n}\left(n\inℕ^∗\right)\)
CMR:1/5S12+1/52S22+.
..+1/5nSn2<1/4