Question

tính tổng \(1^1\)\(+2^2\)\(+3^2\)\(+...\)+\(9^2\)+10

Answer 1

A = 1²+2²+3²+...+9²+10²

A = 1.1+2.2 +3.3 +...+9.9+10.10

A = 1+ 2 ( `1+1) +3 . ( 2+1)+...+ 10 ( 9+1)

A = 1+1.2+ 2+2.3 + 3+...  + 9 . 10+10

A = ( 1.2 + 2.3 +...+9.10) + (1+2+3 +...+10)

Đặt B = 1.2 + 2.3 +...+9.10

=> 3B = 1.2.3 + 2.3 .3 ... 9 .10 . 3

= 9.10.11

= 90 . 11 = 990

=> B = 990 :3 = 330 

=> A = 330 + (1+2+3 +...+10)

=> A = 330 + 55 ( theo t/c)

=> A = 385

Vậy A =385

Answer 2

Sửa đề: A=1^2+2^2+...+9^2+10^2

=10(10+1)*21/6=385