あ 10個月以前 ∑k^2=n/6・(n+1)(2n+1) ∑k(k+1)(k+2) =1/4∑{k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)} f(k)=(k-1)k(k+1)(k+2)とすれば、 ∑k(k+1)(k+2) =1/4∑{f(k+1)-f(k)} =1/4{f(n+1)-f(1)} =n/4・(n+1)(n+2)(n+3) ∴∑(k^3+3k^2+2k)=n/4・(n+1)(n+2)(n+3) ∴∑k^3 =n/4・(n+1)(n+2)(n+3)-3∑k^2-2∑k =n^2/4・(n+1)^2 =(n/2・(n+1))^2 証明終了 留言