Question

Tính tổng 

1^2+2^2+3^2+4^2+...+100^2

Answer 1

Ta có:

1²+2²+3²+4²+...+100²

=1x1 +2x2+3x3+4x4+...+100x100

=1x(1+0)+2x(1+1)+3x(2+1)+4x(3+1)...+100x(99+1)

=1.0+1+1x2+2+2x3+3+3x4+4...+99x100+100

=(1x2+2x3+3x4+...+99x100)+(1+2+3+4+...+100)

=(1x2x3+2x3x3+3x4x3+...+99x100x3):3+(100+1)x100:2

=[(1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+...+99x100x(101-98)]:3+5050

=(1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+...+99x100x101):3 +5050

=99x100x101:3+5050

=333300+5050

=338350

Vậy 1²+2²+3²+...+100²=338350