Lời giải:
$B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}$
$2B=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}$
Trừ theo vế:
$2B-B=1-\frac{1}{2^{2016}}$
$B=1-\frac{1}{2^{2016}}< 1$ (đpcm)
Bài 1: Chứng minh rằng: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2: Cho \(n\in N;n>1\). Chứng minh rằng: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}\notin N\)
Bài 1:So sánh Avà B biết rằng:
A=\(\frac{10^{15}+1}{10^{16}+1};\) B=\(\frac{10^{16}+1}{10^{17}+1}\)
A=\(\frac{3}{8^3}+\frac{7}{8^4}\); B=\(\frac{7}{8^3}+\frac{3}{8^4}\)
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.......+\frac{1}{19}+\frac{1}{20};\) B=\(\frac{1}{2}\)
Bài 2:Dạng tính tổng đặc biệt:
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)
\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+......+\frac{3^2}{340}\)
\(D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^8}\)
\(E=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{99}\right)\)
Bài 3:Dạng chứng minh:
\(A=1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{99}.\)Chứng minh rằng A chia hết cho 100
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\).Chứng minh rằng A>\(\frac{4}{3}\)
Bài 1 : Cho N =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Hãy chứng minh rằng N<1
Bài 1:
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}.\)Chứng minh rằng \(A⋮100\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{2}{13}+...+\frac{1}{70}.\)Chứng minh rằng \(A>\frac{4}{3}\)
Bài 2:Tính \(\frac{A}{B}\)
\(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\) ;\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{9.10}\) ;\(B=\frac{1}{6.10}+\frac{1}{7.9}+\frac{1}{8.8}+\frac{1}{9.7}+\frac{1}{10.6}\)
giúp mk với các nhà toán thông thái à!
Chứng Minh Rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng: a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Nhanh lên nhé! Mk đang cần gấp.
chứng minh rằng
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}\)<1
a) Chứng minh: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\)
b) Chứng minh: \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
chứng minh rằng
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\) bé hơn \(\frac{1}{4}\)
