Thu gọn biểu thức:A=$1^{2}+2^{2}+3^{2}+...+n^{2}$
$A=\frac{n(n+1)(2n+1)}{6}$ dùng quy nạp chứng minh
mak thôi click "V" cho Jin nhờ –  ๖ۣۜJinღ๖ۣۜKaido 19-03-16 07:22 PM
còn cách c/m kia mak dài :3 lm biếng –  ๖ۣۜJinღ๖ۣۜKaido 19-03-16 07:22 PM
ko còn cách nào khác sao –  Lionel Messi 19-03-16 07:20 PM
$A=1^2+2^2+3^2+4^2..+n^2=1+(1+1).2+(1+2).3+(1+3).4+...+(1+n-1).n$
$=1+(2+1.2)+(3+2.3)+(4+3.4)...+n+(n-1)n$
$=(1+2+3+..+n)+[1.2+2.3+3.4+4.5+...(n-1)n]$
Ta có $1+2+3+...+n=\frac{(n+1).n}{2}$
$B=1.2+2.3+3.4+..+(n-1).n$
$\Rightarrow 3B=1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+n.(n-1)[(n+1)-(n-2)]$
  $3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+(n-1)n(n+1)-(n-2)(n-1)n$
  $3B=(n-1)n(n+1)\Rightarrow B=\frac{(n-1)n(n+1)}{3}$
$\Rightarrow A=\frac{n(n+1)}{2}+\frac{(n-1)n(n+1)}{3}=\frac{3n(n+1)+2(n-1)n(n+1)}{6}=\frac{n(n+1)(2n+1)}{6}$

$A=1+(1+1).2+(2+1).3+...+(n-1+1).n$
$\Leftrightarrow A=1+1.2+2+2.3+3+...+(n-1).n+n$
$\Leftrightarrow A=(1+2+3+...+n)+[1.2+2.3+3.4+...+(n-1).n]$
Đặt $B=1.2+2.3+3.4+...(n-1).n$
$\Leftrightarrow 3B=1.2.3+2.3.(4-1)+3.4.(5-2)+....+(n-1)n[(n+1)-(n-2)]$
$\Leftrightarrow 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+(n-1)n(n+1)-(n-2)(n-1)n$
$\Leftrightarrow 3B=(n-1)n(n+1)\Leftrightarrow B=\frac{(n-1)n(n+1)}{3}$
Do đó $A=\frac{n(n+1)}{2}+\frac{(n-1)n(n+1)}{3}=\frac{n(n+1)(2n+1)}{6}$
nó nek :3 –  ๖ۣۜJinღ๖ۣۜKaido 19-03-16 07:36 PM
có phải cách này ko –  Lionel Messi 19-03-16 07:31 PM

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