(\(\frac{1}{2^2}-1\)).(\(\frac{1}{3^2}-1\)).(\(\frac{1}{4^2}-1\)).......(\(\frac{1}{100^2}-1\))
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63 = (2.3)3 = 23.33 < 33.33 = 36
Vậy 36>63
162 = (24)2 = 28 < 29
Vậy 162 < 29
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
ta có n+7=n-1+8
=>8 chia hết cho n-1
=>n-1 thuộc U(8)=+-1;+-2;+-4;+-8
=>n-1=1=>n=2
n-1=-1=>n=0
n-1=2=>n=3
n-1=-2=>n=-1
n-1=4=>n=5
n-1=-4=>n=-3
n-1=8=>n=9
n-1=-8=>n=-7
vậy n=0;3;-1;5;-3;9;-7;2
2S=2+2^2+2^3+...+2^101
2S-S=2^101-1
S=2^101-2<2^101
hok tốt
\(S=1+2+2^2+\cdot\cdot\cdot+2^{100}\)
\(\Rightarrow2S=2+2^2+2^3+\cdot\cdot\cdot+2^{101}\)
\(\Rightarrow2S-S=\left(2+\cdot\cdot+2^{101}\right)-\left(1+\cdot\cdot\cdot+2^{100}\right)\)
\(\Rightarrow S=2^{101}-1\)<\(2^{101}\)
\(\Rightarrow S\)<\(2^{101}\)
Ta có:\(-\left(n-1\right)\left(n+1\right)=-\left(n^2+n-n-1\right)=-n^2+1=1-n^2\)
Xét:\(A=...\)
\(=\frac{1-2^2}{2^2}\cdot\frac{1-3^2}{3^2}\cdot...\cdot\frac{1-100^2}{100^2}\)
Mà A có:99 số hạng => số lẻ,nên:
\(A=-\left(\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{100^2-1}{100^2}\right)\)
\(=-\left(\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{99\cdot101}{100^2}\right)\)
\(=-\left(\frac{1}{2}\cdot\frac{101}{100}\right)=-\frac{101}{200}\)
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=\(-\frac{3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-9999}{10000}\)
vì có 50 thừa số nên A= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
=>A=\(\frac{3.8.15.....9999}{4.9.16.....10000}\)
A=\(\frac{\left(1.3\right)\left(2.4\right)\left(3.5\right).....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)
A=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
A=\(\frac{1.101}{100.2}\)
A=\(\frac{101}{200}\)