1x2+2x3+3x4+...+n(n+1)=?
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1x2+2x3+3x4+...+n(n+1)=?1x2+2x3+3x4+...+n(n+1)=?1x2+2x3+3x4+...+n(n+1)=?1x2+2x3+3x4+...+n(n+1)=1^2+1
1x2+2x3+3x4+...+n(n+1)=?
1x2+2x3+3x4+...+n(n+1)=?
1x2+2x3+3x4+...+n(n+1)=?
1x2+2x3+3x4+...+n(n+1)
=1^2+1+2^2+2+3^2+3+...+n^2+n
=1+2+...+n+(1^2+2^2+...+n^2)
=(1+n)n/2+n(n+1)(2n+1)/6
=n(n+1)/2*(1+(2n+1)/3)
=n(n+1)(2n+5)/6
---看通项,再分解,利用公式!
1*2=1*3(1*2*3-0*1*2)
2*3=1/3(2*3*4-1*2*3)
...............
n*(n+1)=1/3[n*(n+1)*(n+2)-(n-1)*n*(n+1)]
所以
1x2+2x3+3x4+···+n(n+1)
=1/3[n*(n+1)*(n+2)-0*1*2]
=1/3n*(n+1)*(n+2)
1x2+2x3+3x4+...+n(n+1)
=(1²+1)+(2²+2)+(3²+3)+...+(n²+n)
=(1²+2²+....n²)+(1+2+...+n)
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