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A = 22 + 42 + 62 + 82 + ... + 202
A = 22.(12 + 22 + 32 + 42 + ... + 102)
A = 4.385
A = 1540
A = 22 + 42 + 62 + 82 + ... + 202
A = 22 . (12 + 22 + 32 + 42 + ... + 102)
A = 4 . 385
A = 1540
A = 22 + 42 + 62 + ... + 202 = 1540
A = 22 . ( 12 + 22 + 32 + ... + 102 ) = 1540
\(\Rightarrow\)12 + 22 + 32 + ... + 102 = 1540 : 22 = 385
Vậy B = 385
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
22+42+62+...+202=12.22+22.22+32.22+....+102.22
=(12+22+32+...+102).4=385.4=1540
A=4+16+36+64+100+144+196+256+324+400
A=20+100+100+340+580+400
A=320+920+400
A=1240+400
A=1640
\(A=2^2+4^2+6^2+...+20^2.\)
\(=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(=2^2\left[1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+10\left(11-1\right)\right]\)
\(=2^2\left[\left(1.2+2.3+3.4+...+10.11\right)-\left(1+2+3+...+10\right)\right]\)
\(=2^2\left[\frac{1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+10.11.\left(12-9\right)}{3}-\frac{10\left(10+1\right)}{2}\right]\)
\(=2^2\left[\frac{1.2.3-0.1.2+2.3.4-1.2.3+...+10.11.12-9.10.11}{3}-55\right]\)
\(=2^2\left(\frac{10.11.12}{3}-55\right)=2^2.275=1100\)