Đáp án: $A=\dfrac{(x-1).x.(x+1)}{3}+\dfrac{x(x+1)}{2}$
Giải thích các bước giải:
Ta có:
$A=1^2+2^2+...+x^2$
$\to A=1.1+2.2+...+x.x$
$\to A=(0+1).1+(1+1).2+...+(x-1+1).x$
$\to A=(0.1+1.2+...+(x-1).x)+(1+2+...+x)$
$\to A=(1.2+...+(x-1).x)+\dfrac{x(x+1)}{2}$
Mà $B=1.2+2.3+...+(x-1).x$
$\to 3B=1.2.3+2.3.3+...+(x-1).x.3$
$\to 3B=1.2.(3-0)+2.3.(4-1)+...+(x-1).x.(x+1-(x-2))$
$\to 3B=-0.1.2+1.2.3-1.2.3+2.34.+...-(x-2).(x-1).x+(x-1).x.(x+1)$
$\to 3B=(x-1).x(x+1)$
$\to B=\dfrac{(x-1).x.(x+1)}{3}$
$\to A=\dfrac{(x-1).x.(x+1)}{3}+\dfrac{x(x+1)}{2}$
