so sánh \(\frac{1}{16}\) với tổng A gồm 11 số hạng A=\(\frac{1}{5^2}\)+\(\frac{2}{5^3}\)+...+\(\frac{5}{5^{n+1}}\)+....+\(\frac{11}{5^{12}}\)
Cho \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{\text{n}}{5^{\text{n}-2}}+...+\frac{11}{5^{12}}\) với \(\text{n}\in\text{N }\).CMR:\(A< \frac{1}{16}\)
Cho A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)với n\(\inℕ\).Chứng minh rằng A<\(\frac{1}{16}\)
Giúp mình với, hiện đang cần gấp lắm.
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
Cho A = \(\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+.......+\frac{n}{5^{n+1}}+.......+\frac{11}{5^{12}}\) với n \(\in\) N. Chứng minh rằng A < \(\frac{1}{16}\)
Giúp mk vs
Cho A= \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+..............+\frac{11}{5^{12}}\)so sanh A voi \(\frac{1}{16}\)
Cho \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
Chứng minh A<\(\frac{1}{16}\)
Cho \(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)với \(n\in N.\)Chứng minh rằng \(A<\frac{1}{16}\)
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{n}{5^{n+1}}+\frac{11}{5^{12}}\)với \(n\in N.\) Chứng minh rằng \(A<\frac{1}{16}\)
Cho A =\(\frac{1}{^{5^2}}\)+\(\frac{2}{5^3}\)+\(\frac{3}{5^4}\)+...+\(\frac{n}{5^{n+1}}\)+...+\(\frac{11}{5^{12}}\)với n thuộc N chứng minh rằng A<\(\frac{1}{16}\)
tính b
b=\(\frac{0,275-0,5+\frac{3}{11}}{0,625+0,5+\frac{5}{11}}:\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
So sánh
A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^9}+.....+\frac{1}{3^{33}}với\frac{1}{2}\)