HT
2
K
Khách
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LV
0
NT
1
BT
19 tháng 4 2018
\(A=1-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt: \(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2B-B=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> \(B=2-\frac{1}{2^{10}}\)
=> \(A=1-B=1-2+\frac{1}{2^{10}}\)
=> \(A=\frac{1}{2^{10}}-1\)
PM
1
LS
9 tháng 7 2018
\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot..\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\left(\frac{1}{2}\cdot\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}\cdot\frac{1}{3}-1\right)\cdot...\cdot\left(\frac{1}{10}\cdot\frac{1}{10}-1\right)\)
\(=\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)
\(=\frac{\left(-1\right).\left(-3\right)}{2.2}\cdot\frac{\left(-2\right).\left(-4\right)}{3.3}\cdot...\cdot\frac{\left(-9\right).\left(-11\right)}{10.10}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(-9\right)}{2.3....10}\cdot\frac{\left(-3\right).\left(-4\right)....\left(-11\right)}{2.3.....10}\)
\(=\frac{-1}{10}\cdot\frac{-11}{2}=\frac{-11}{20}\)
DT
3
31 tháng 8 2015
= \(\frac{20.21:2+2870}{2}=\frac{210+2870}{2}=\frac{3080}{2}=1540\)
31 tháng 8 2015
\(=\frac{1\left(1+1\right)}{2}+\frac{2\left(2+1\right)}{2}+\frac{3\left(3+1\right)}{2}+...+\frac{20\left(20+1\right)}{2}\)
\(=\frac{1+1+2.2+2+3.3+3+...+20.20+20}{2}\)
\(=\frac{\left(1+...+20\right)+\left(1.1+2.2+3.3+...+20.20\right)}{2}\)
Tính tiếp đi
22 tháng 2 2022
a: \(1^3+2^3+3^3+4^3+5^3=225\)
\(\left(1+2+3+4+5\right)^2=15^2=225\)
Do đó: \(1^3+2^3+3^3+4^3+5^3=\left(1+2+3+4+5\right)^2\)
b: \(1^3+2^3+...+10^3=3025\)
\(\left(1+2+3+...+10\right)^2=55^2=3025\)
Do đó: \(1^3+2^3+...+10^3=\left(1+2+3+...+10\right)^2\)
HT
1
KL
1
7 tháng 2 2018
N=\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{10^2}\right)\)
= \(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{8.10}{9^2}\)
= \(\frac{1.2.3...8}{2.3...9}.\frac{3...8.9.10}{2.3...9}\)
=\(\frac{1}{9}.\frac{10}{2}\)
= \(\frac{1}{9}.5=\frac{5}{9}\)
HP
0
\(\left(1-\frac{1}{2}^2\right).\left(1-\frac{1}{3}^2\right).....\left(1-\frac{1}{10}^2\right)\)
=\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{100}\right)\)
= \(\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
= \(\frac{1.3.2.4.....9.11}{2.2.3.3.....10.10}\)
= \(\frac{1.11}{2.10}\)
= \(\frac{11}{20}\)
11/20