tinh S=1mu2+2mu2+.....+10mu2
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S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385
= 1540
= (1x2)^2 (2x2)^2 (3x2)^2 (4x2)^2 ..... (9x2)^2 (10x2)^2
= 1^2 x 2^2 2^2 x 2^2 3^2 x 2^2 4^2 x 2^2 ..... 9^2 x 2^2 10^2 x 2^2
= (1^2 2^2 3^2 4^2 ..... 9^2 10^2) x 2^2
= 385 x 2^2 = 385 x 4 = 1540
\(\left[1-\frac{1}{2^2}\right]\left[1-\frac{1}{3^2}\right]\left[1-\frac{1}{4^2}\right]...\left[1-\frac{1}{10^2}\right]\)
\(=\left[1-\frac{1}{4}\right]\left[1-\frac{1}{9}\right]\left[1-\frac{1}{16}\right]...\left[1-\frac{1}{100}\right]\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}\)
Tự tính :v
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{20}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{19}}\)
=> \(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> \(S=1-\frac{1}{2^{20}}\)
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)
\(10^2\cdot10^3\cdot10^4\cdot10^5\cdot10^6\)
\(=10^{2+3+4+5+6}\)\
\(=10^{20}\)
Trả lời:
102 . 103 . 104 . 105 . 106
= 102+3+4+5+6
= 1020
Hok tốt!
Đạt A = 2 + 2 2 + ... + 2 500
2A = 2 2 + 2 3 + .... + 2 501
2A - A = ( 2 2 + 2 3 + .... + 2 501 )
- ( 2 + 2 2 + ... + 2 500 )
A = 2 501 - 2
Dat A =1+2+22+23+...+250
=> 2A=2+22+23+24+...+251
=> 2A-A= 2+22+23+24+...+250 - ( 1+2+22+23+...+250 )
=> A=251-1
22s=2+22+...+22020
4S-S=(2+22+...+22020)-(1+2+22+....+22018)
3S=22020-1
S=(22020-1):3
S = 12 + 22 + ... + 102
S = 1 + 4 + ... + 100
S = 385
2S=2+22+23+....+210+211
2S-S=211-1
S=211-1