1×2+2×3+...+n(n+1)?

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1×2+2×3+...+n(n+1)?
1×2+2×3+...+n(n+1)?

1×2+2×3+...+n(n+1)?
k(k+1)=1/3*k(k+1)[(k+2)-(k-1)]=1/3[k(k+1)(k+2)-(k-1)k(k+1)]所以原式=1/3*[1*2*3-0*1*2+2*3*4-1*2*3+.....+n(n+1)(n+2)-(n-1)n(n+1)]=1/3*n(n+1)(n+2)