k平方的前n项和求解过程

2024-09-06 来源：易榕旅网

k_=K(K+1)(K+2)/3-(K-1)K(K+1)/3+k(k+1)/2-K(K-1)/2
∴1_+2_+...+k_=1*2*3/3-0*1*2/3+1*2/2-1*0/2+.....+k(K+1)(K+2)/3-(k-1)K(K+1)/3+K(K+1)/2-K(K-1)/2=k(k+1)(2k+1)/6
即:1_+2_+3_+.....+K_=K(K+1)(2K+1)/6

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