..., 16, 32, 64, 4096.

1; 1; 2; 2; 4; 16; 32; 64; 4096

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\)

= \(\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}...\dfrac{10^2-1}{10^2}\)

= \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{9.11}{10^2}\)

= \(\dfrac{\left(1.2.3...9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)

= \(\dfrac{1.11}{10.10}=\dfrac{11}{100}\)

quá hay

Sửa đề: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)

Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)

\(=\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{100}+1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}\)

\(=-\frac{1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)

\(C=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+2016}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{\dfrac{\left(2016+1\right).2016}{2}}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{2033136}\right)\)

\(=\dfrac{2}{3}.\dfrac{5}{6}...\dfrac{2033135}{2033136}\)

\(=\dfrac{4}{6}.\dfrac{10}{12}...\dfrac{4066270}{4066272}\)

\(=\left(\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2015}{2016}\right).\left(\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{2018}{2017}\right)\)

\(=\dfrac{1}{2016}.\dfrac{2018}{3}=\dfrac{1009}{3024}\)

cảm ơn cậu

\(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)

\(=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)

\(=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Do đó, ta có: \(1-\frac{1}{1+2}=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}=\frac{4\cdot1}{2\cdot3}\)

\(1-\frac{1}{1+2+3}=\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}=\frac{5\cdot2}{3\cdot4}\)

...

\(1-\frac{1}{1+2+\cdots+n}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Do đó: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+3+\cdots+n}\right)\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(=\frac{4\cdot5\cdot\ldots\cdot\left(n+2\right)}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}\cdot\frac{1\cdot2\cdot\ldots\cdot\left(n-1\right)}{2\cdot3\cdot\ldots\cdot n}=\frac{n+2}{3}\cdot\frac{1}{n}=\frac{n+2}{3n}\)

A =             1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)

A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) +  \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)

A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)

A \(\times\)( 2-1) = \(\dfrac{255}{128}\)

A = \(\dfrac{255}{128}\)

Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T

\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)

\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)

\(T=2+0+0+...-\dfrac{1}{128}\)

\(T=\dfrac{256}{128}-\dfrac{1}{128}\)

\(T=\dfrac{255}{128}\)

A=[(1-2²)/2²][(1-3²)/3²]...[(1-2015²)/2015²]

A=[(1+2)(1-2)/2²][(1-3)(1+3)/3²]...[(1-2015)(1+2015)/2015²]

=[(-1).3/2.2][(-2).4/3.3]...[-2014.2016/2015.2015]

=[(-1)(-2)(-3)...(-2013)(-2014).3.4.5...2015]/(2.2.3.3.4.4....2015.2015)

=[2(-3)...(-2014)]/(2.2.3.4.5....2015)

=(-3)(-4)...(-2014)/2.3.4.5....2015

=[-(3.4.5.6....2014)]/(2.3.4...2015)

=-1/1.2015=-1/2015

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)

\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)

\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)

\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)

\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).