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Question

Thu gọn biểu thức:A=$1^{2}+2^{2}+3^{2}+...+n^{2}$

๖ۣۜJinღ๖ۣۜKaido · Answer 1 · 3/19/2016 7:36:24 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 $1+2+3+...+n=\frac{(n+1).n}{2}$

$B=1.2+2.3+3.4+..+(n-1).n$

$\Rightarrow 3B=1.2.(3-0)+2.3(4-1)+3.4(5-2)+...+n.(n-1)[(n+1)-(n-2)]$

$3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+(n-1)n(n+1)-(n-2)(n-1)n$

$3B=(n-1)n(n+1)\Rightarrow B=\frac{(n-1)n(n+1)}{3}$

$\Rightarrow A=\frac{n(n+1)}{2}+\frac{(n-1)n(n+1)}{3}=\frac{3n(n+1)+2(n-1)n(n+1)}{6}=\frac{n(n+1)(2n+1)}{6}$

Trả lời 19-03-16 07:36 PM

๖ۣۜJinღ๖ۣۜKaido
1

10M 1M

๖ۣۜJinღ๖ۣۜKaido · Answer 2 · 3/19/2016 7:02:07 PM

$A=\frac{n(n+1)(2n+1)}{6}$ dùng quy nạp chứng minh

Trả lời 19-03-16 07:02 PM

๖ۣۜJinღ๖ۣۜKaido
1

10M 1M

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

Lionel Messi · Answer 3 · 3/19/2016 7:30:59 PM

$A=1+(1+1).2+(2+1).3+...+(n-1+1).n$

$\Leftrightarrow A=1+1.2+2+2.3+3+...+(n-1).n+n$

$\Leftrightarrow A=(1+2+3+...+n)+[1.2+2.3+3.4+...+(n-1).n]$

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 $A=\frac{n(n+1)}{2}+\frac{(n-1)n(n+1)}{3}=\frac{n(n+1)(2n+1)}{6}$

Trả lời 19-03-16 07:30 PM

Lionel Messi
1

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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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