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Đặt T = 12 + 22 + ... + 102 = 385
=> T x 22 = 12. 22 + 22. 22 + ... + 102.22 = 385. 22
=> T x 22 = (1.2)2 + (2. 2)2 + ... + (10.2)2 = 385. 22
=> T x 22 = (2)2 + (4)2 + ... + (20)2 = 385. 22
=> T x 22 = S = 385. 22
=> S = 385 x 4
olm duyệt
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi tổng đầu là A
22A=(1.2)2+(2.2)2+...+(2.10)2
4A = 22+42+62+..+202=> 4A = S = 385 .4 = 1460
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
S=22+42+62+...+202
S=(2*1)2+(2*2)2+(2*3)2+...+(2*10)2
=22*12+22*22+...+22*102=22(12+22+32+...+102)=4*385=1540
Vậy S=1540
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^2\left(1^2+2^2+3^2+...+10^2\right)=2^2+4^2+6^2+...+20^2=S\)
=> \(S=2^2.385=1540\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2^2+4^2+6^2+....+20^2\)
\(=\left(1.2\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+...+\left(2.10\right)^2\)
\(=1^2.2^2+2^2.2^2+2^2.3^2+....+2^2.10^2\)
\(=2^2.\left(1^2+2^2+3^2+....+10^2\right)\)
Mà \(1^2+2^2+3^2+...+10^2=385\)
Nên \(S=2^2.385=4.385=1540\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)
\(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)
\(=6^2.1771=36.1771=63756\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2^2+4^2+.....+20^2\)
\(S=1^2.2^2+2^2.2^2+.......+10^2.2^2\)
\(S=2^2.\left(1^2+2^2+.....+10^2\right)\)
\(S=4.385=1540\) (đề bài)
1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 +81 + 100 + 121 = 506
506