Σ[k=1,n](2k-1)^3などの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n (2k-1)=n^2\\
&(2)  \displaystyle\sum_{k=1}^n (2k-1)^2=\frac{1}{3}n(4n^2-1)\\
&(3)  \displaystyle\sum_{k=1}^n (2k-1)^3=n^2(2n^2-1)\\
&(4)  \displaystyle\sum_{k=1}^n (mk-1)=\frac{1}{2}n\{m(n+1)-2\}\\
&(5)  \displaystyle\sum_{k=1}^n (mk-1)^2=\frac{1}{6}n\{m^2(n+1)(2n+1)-6m(n+1)+6\}\\
&(6)  \displaystyle\sum_{k=1}^n (mk-1)^3=\frac{1}{4}n\{m^3n(n+1)^2-2m^2(n+1)(2n+1)+6m(n+1)-4\}
\end{alignat}










<証明>

次の等式を用います。(詳細はこちらです。)
\begin{alignat}{2}
&(A)  \displaystyle\sum_{k=1}^n k=\frac{1}{2}n(n+1)\\
&(B)  \displaystyle\sum_{k=1}^n k^2=\frac{1}{6}n(n+1)(2n+1)\\
&(C)  \displaystyle\sum_{k=1}^n k^3=\frac{1}{4}n^2(n+1)^2\\
\end{alignat}





\begin{alignat}{2}
(1)  \displaystyle\sum_{k=1}^n (2k-1)&=2\displaystyle\sum_{k=1}^n k-\displaystyle\sum_{k=1}^n 1=2 \cdot \frac{1}{2}n(n+1)-n\\
&=n(n+1)-n=n^2-n+n=n^2\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (2k-1)=n^2$$







\begin{alignat}{2}
(2)  \displaystyle\sum_{k=1}^n (2k-1)^2&=\displaystyle\sum_{k=1}^n (4k^2-4k+1)=4\displaystyle\sum_{k=1}^n k^2-4\displaystyle\sum_{k=1}^n k+\displaystyle\sum_{k=1}^n 1\\
&=4 \cdot \frac{1}{6}n(n+1)(2n+1)-4 \cdot \frac{1}{2}n(n+1)+n\\
&=\frac{2}{3}n(n+1)(2n+1)-2n(n+1)+n\\
&=\frac{2}{3}n(2n^2+3n+1)-2n(n+1)+n\\
&=\frac{4}{3}n^3+2n^2+\frac{2}{3}n-2n^2-2n+n\\
&=\frac{4}{3}n^3-\frac{1}{3}n=\frac{1}{3}n(4n^2-1)\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (2k-1)^2=\frac{1}{3}n(4n^2-1)$$







\begin{alignat}{2}
(3)  \displaystyle\sum_{k=1}^n (2k-1)^3&=\displaystyle\sum_{k=1}^n (8k^3-12k^2+6k-1)\\
&=8\displaystyle\sum_{k=1}^n k^3-12 \displaystyle\sum_{k=1}^n k^2+6 \displaystyle\sum_{k=1}^n k-\displaystyle\sum_{k=1}^n 1\\
&=8 \cdot \frac{1}{4}n^2(n+1)^2-12 \cdot \frac{1}{6}n(n+1)(2n+1) +6 \cdot \frac{1}{2}n(n+1)-n\\
&=2n^2(n+1)^2-2n (n+1)(2n+1)+3n(n+1)-n\\
&=2n^2(n^2+2n+1)-2n(2n^2+3n+1)+3n(n+1)-n\\
&=2n^4+4n^3+2n^2-4n^3-6n^2-2n+3n^2+3n-n\\
&=2n^4-n^2=n^2(2n^2-1)\\
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (2k-1)^3=n^2(2n^2-1)$$







\begin{alignat}{2}
(4)  \displaystyle\sum_{k=1}^n (mk-1)&=m\displaystyle\sum_{k=1}^n k-\displaystyle\sum_{k=1}^n 1=m \cdot \frac{1}{2}n(n+1)-n\\
&=\frac{1}{2}n\{(m+1)-2\}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (mk-1)=\frac{1}{2}n\{m(n+1)-2\}$$








\begin{alignat}{2}
(5)  \displaystyle\sum_{k=1}^n (mk-1)^2&=\displaystyle\sum_{k=1}^n (m^2k^2-mk+1)=m^2\displaystyle\sum_{k=1}^n k^2-2m\displaystyle\sum_{k=1}^n k+\displaystyle\sum_{k=1}^n 1\\
&=m^2 \cdot \frac{1}{6}n(n+1)(2n+1)-2m \cdot \frac{1}{2}n(n+1)+n\\
&=\frac{1}{6}m^2n(n+1)(2n+1)-mn(n+1)+n\\
&=\frac{1}{6}n\{m^2(n+1)(2n+1)-6m(n+1)+6\}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (mk-1)^2=\frac{1}{6}n\{m^2(n+1)(2n+1)-6m(n+1)+6\}$$







\begin{alignat}{2}
(6)  \displaystyle\sum_{k=1}^n (mk-1)^3&=\displaystyle\sum_{k=1}^n (m^3k^3-3m^2k^2+3mk-1)\\
&=m^3\displaystyle\sum_{k=1}^n k^3-3m^2 \displaystyle\sum_{k=1}^n k^2+3m \displaystyle\sum_{k=1}^n k-\displaystyle\sum_{k=1}^n 1\\
&=m^3 \cdot \frac{1}{4}n^2(n+1)^2-3m^2 \cdot \frac{1}{6}n(n+1)(2n+1) +3m\cdot \frac{1}{2}n(n+1)-n\\
&=\frac{1}{4}m^3n^2(n+1)^2-\frac{1}{2}m^2n(n+1)(2n+1)+\frac{3}{2}mn(n+1)-n\\
&=\frac{1}{4}n\{m^3n(n+1)^2-2m^2(n+1)(2n+1)+6m(n+1)-4\}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n (mk-1)^3=\frac{1}{4}n\{m^3n(n+1)^2-2m^2(n+1)(2n+1)+6m(n+1)-4\}$$






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