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a/
$A-3=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}-3$
$=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})-3$
$=\frac{2}{2003}-\frac{1}{2004}-\frac{1}{2005}$
$=(\frac{1}{2003}-\frac{1}{2004})+(\frac{1}{2003}-\frac{1}{2005})$
$>0+0=0$
$\Rightarrow A>3$
b/
$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}$
$=1-\frac{1}{2015}<1$
ta có:
\(a=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
\(a=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{9801}+\frac{1}{10000}\)
\(a=\left(\frac{1}{4}+\frac{1}{10000}\right)+\left(\frac{1}{9}+\frac{1}{9801}\right).\left(10000-4:\left(9-4\right)\right)\)
a=\(\frac{1}{10004}.498=\frac{249}{5002}\)
vì:\(\frac{249}{5002}< \frac{3}{4}=>a< \frac{3}{4}\)
a)tính dễ
b)chứng minh nó = quy nạp thôi
n=1 và n=k; n=k+1;... trong trang cá nhân mk lm r` đó bn chịu khó tìm lại
\(A=\left[\frac{1}{2^2}-1\right]\left[\frac{1}{3^2}-1\right]\left[\frac{1}{4^2}-1\right]\cdot...\cdot\left[\frac{1}{100^2}-1\right]\)
\(=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-9999}{100^2}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot...\cdot\frac{-99\cdot101}{100\cdot100}\)
\(=\frac{-1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot...\cdot100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)
Mà \(-\frac{101}{200}< -\frac{1}{2}\)
nên \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}\)
\(A=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}...\frac{-99.101}{100.100}\)
\(A=\frac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(A=-\frac{1}{100}.\frac{101}{2}\)
\(A=-\frac{101}{200}\)
\(\text{Vậy A=}-\frac{101}{200}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=2-\frac{1}{100}< 2\)
Vậy \(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 2\)
=>A<2(đpcm)
Ta có: A = 1 + 1/22+1/32+1/4^2+...+1/100^2 < 1+1/1.2 +1/2.3+1/3.4+...+1/99.100
=> A < 1+(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)
A < 1+(1-1/100)
A < 1+99/100
Vì 1+99/100 < 2 nên A < 2
\(3^{21}=3.3^{20}=3.\left(3^2\right)^{10}=3.9^{10}\)
\(2^{31}=2.2^{30}=2.\left(2^3\right)^{10}=2.8^{10}\)
Thấy: 3 > 2 và 910 > 810
Nên \(3^{21}>2^{31}\)
Bài 2:
\(A=1+2+2^2+.....+2^{100}\)
\(2A=2+2^2+.......+2^{101}\)
\(2A-A=\left(2-2\right)+\left(2^2-2^2\right)+......+2^{101}-1\)
Vậy A = 2101 - 1