解:
设第n项为an
an=n(n+1)(n+2)=n^3+3n^2+2n
1×2×3+2×3×4+...+10×11×12
=(1^3+2^3+...+10^3)+3(1^2+2^2+...+10^2)+2(1+2+...+10)
=[10(10+1)/2]^2+3×10×(10+1)(20+1)/6+2×10×11/2
=3025+1155+110
=4290
用到的公式:
1^3+2^3+...+n^3=[n(n+1)/2]^2
1^2+2^2+...+n^2=n(n+1)(2n+1)/6
1+2+3+...+n=n(n+1)/2
1×2×3+2×3×4+3×4×5+……+10×11×12
=10×(10+1)×(10+2)×(10+3)/4
=4290
∑▒〖k(k+10(k+2)〗=∑▒〖(k^3+3k^2+2k)〗,∑_1^n▒n^3 =[n(n+1)/2]^2,∑_1^n▒n^2 =n(n+1)(2n+1)/6,∑_1^n▒n=n(n+1)/2
所以原式=∑_1^10▒〖(k^3+3k^2+2k)〗=[(10*(10+1))/2]^2+3*(10*(10+1)*(2*10+1))/6+10*(10+1)=4290