Tính: A = $1^{2}$ + $2^{2}$ + $3^{2}$ +…….+ $x^{2}$

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1. 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 $B=1.2+2.3+…+(x-1).x$

   $\to 3B=1.2.3+2.3.3+…+(x-1).x.3$

   $\to 3B=1.2.(3-0)+2.3.(4-1)+…+(x-1).x.(x+1-(x-2))$

   $\to 3B=-0.1.2+1.2.3-1.2.3+2.34.+…-(x-2).(x-1).x+(x-1).x.(x+1)$

   $\to 3B=(x-1).x(x+1)$

   $\to B=\dfrac{(x-1).x.(x+1)}{3}$

   $\to A=\dfrac{(x-1).x.(x+1)}{3}+\dfrac{x(x+1)}{2}$

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