数列求和问题,好难好难好难 1×4+2×7+3×10+…+(n+1)(3n+4)=

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数列求和问题,好难好难好难 1×4+2×7+3×10+…+(n+1)(3n+4)=
数列求和问题,好难好难好难 1×4+2×7+3×10+…+(n+1)(3n+4)=

数列求和问题,好难好难好难 1×4+2×7+3×10+…+(n+1)(3n+4)=
an=3n^2+7n+4
1^2+2^2+3^2+……+n^2=1/6*n(n+1)(2n+1)
1+2+3+……+n=n(n+1)/2
1×4+2×7+3×10+…+(n+1)(3n+4)
=3[1^2+2^2+3^2+……+n^2]+7[1+2+3+……+n]+4n
=n(n+1)(2n+1)/2+7n(n+1)/2+4n
=n(n+1)(2n+1+7)/2+4n
=n(n+1)(n+4)+4n

1×4+2×7+3×10+…+(n+1)(3n+4)=n(n+1)的平方

给你一种方法:
(n+1)(3n+4) = ((n+1)^3-n^3)/3+6n+11/3