Cho: \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}\)
\(CMR:B< 1\)
K
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15 tháng 8 2016
B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
=> 2B=\(2\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]\)
=\(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{98}\)
=>2B-B=\(\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{98}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\right]\)
=>B=\(1-\left(\frac{1}{2}\right)^{99}< 1\)
=> B<1
KB
26 tháng 3 2019
1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
A= \(\frac{100}{2}\)
A=50
2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)
B= \(\frac{1}{99}\)
HN
26 tháng 3 2019
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)
\(=\frac{100}{2}\)
\(=50\)
\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)
\(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)
\(=-\frac{1}{99}\)
S
2 tháng 12 2019
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}=\frac{2^{99}-1}{2^{99}}\)
Ta có: \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow B=1-\left(\frac{1}{2}\right)^{99}< 1\)