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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: \(1^3+2^3+...+10^3=\left(1+2+3+...+10\right)^2\)
\(A=1-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt: \(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2B-B=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> \(B=2-\frac{1}{2^{10}}\)
=> \(A=1-B=1-2+\frac{1}{2^{10}}\)
=> \(A=\frac{1}{2^{10}}-1\)
\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot..\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\left(\frac{1}{2}\cdot\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}\cdot\frac{1}{3}-1\right)\cdot...\cdot\left(\frac{1}{10}\cdot\frac{1}{10}-1\right)\)
\(=\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)
\(=\frac{\left(-1\right).\left(-3\right)}{2.2}\cdot\frac{\left(-2\right).\left(-4\right)}{3.3}\cdot...\cdot\frac{\left(-9\right).\left(-11\right)}{10.10}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(-9\right)}{2.3....10}\cdot\frac{\left(-3\right).\left(-4\right)....\left(-11\right)}{2.3.....10}\)
\(=\frac{-1}{10}\cdot\frac{-11}{2}=\frac{-11}{20}\)
= \(\frac{20.21:2+2870}{2}=\frac{210+2870}{2}=\frac{3080}{2}=1540\)
\(=\frac{1\left(1+1\right)}{2}+\frac{2\left(2+1\right)}{2}+\frac{3\left(3+1\right)}{2}+...+\frac{20\left(20+1\right)}{2}\)
\(=\frac{1+1+2.2+2+3.3+3+...+20.20+20}{2}\)
\(=\frac{\left(1+...+20\right)+\left(1.1+2.2+3.3+...+20.20\right)}{2}\)
Tính tiếp đi
\(\left(1-\frac{1}{2}^2\right).\left(1-\frac{1}{3}^2\right).....\left(1-\frac{1}{10}^2\right)\)
=\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{100}\right)\)
= \(\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
= \(\frac{1.3.2.4.....9.11}{2.2.3.3.....10.10}\)
= \(\frac{1.11}{2.10}\)
= \(\frac{11}{20}\)
Bài 1
a/
\(A=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+10\left(11-1\right)=\)
\(=\left(1.2+2.3+3.4+...+10.11\right)-\left(1+2+3+...+10\right)=\)
Đặt \(B=1.2+2.3+3.4+...+10.11\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+10.11.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+10.11.\left(12-9\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-9.10.11+10.11.12=\)
\(=10.11.12\Rightarrow B=\frac{10.11.12}{3}=4.10.11\)
\(\Rightarrow A=B-\left(1+2+3+...+10\right)=4.10.11+\frac{10.\left(1+10\right)}{2}=\)
\(=4.10.11+5.11=11.\left(4.10+5\right)=11.45=495\)
b/
\(B=5^2\left(1+2^2+3^2+...+10^2\right)=25.495=12375\)
Bài 2
Số số hạng của M \(=\frac{2n-1-1}{2}+1=n\)
\(M=\frac{n\left[1+\left(2n-1\right)\right]}{2}=n^2\)là số chính phương
Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
Lấy \(2A-A\), ta có:
\(2A-A=A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2^9}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=\left(1-\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{2^9}-\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Vậy \(A=\frac{1023}{1024}\)