Σ計算の証明
1. 定数 c
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&\sum_{k=1}^{n}c = c + c + c + ... = nc（c を n 個足す）\\
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2. 変数 k
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&\begin{array}{rll}
&1 &+ 2 &+ 3 &+ ... &+ n&\\
+) &n &+ n-1 &+ n-2 &+ ... &+ 1&\\
&--- &---- &---- &--- &----&\\
&(n+1) &+ (n+1) &+ (n+1) &+ ... &+ (n+1) &= (n+1) \cdot n \\
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&2\sum_{k=1}^{n}k = n(n + 1)\\
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&\sum_{k=1}^{n}k = \frac{n(n + 1)}{2}\\
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3. 変数 k^2
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&(k+1)^3 - k^3 = 3k^2 + 3k + 1 \because (k+1)^3 = k^3 + 3k^2 + 3k + 1 \\
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&\begin{array}{rll}
&2^3 &-1^3 &= 3\cdot1^2 &+ 3\cdot1 &+ 1\\
&3^3 &-2^3 &= 3\cdot2^2 &+ 3\cdot2 &+ 1\\
&4^3 &-3^3 &= 3\cdot3^2 &+ 3\cdot3 &+ 1\\
&&:&:\\
+) &(n+1)^3 &-n^3 &= 3\cdot n^2 &+ 3\cdot n &+ 1\\
&--- &---- &---- &--- &----&\\
&(n+1)^3 &-1^3 &= 3\sum_{k=1}^{n}k^2 &+ 3\sum_{k=1}^{n}k &+ n\\
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&\begin{array}{rll}
&3\sum_{k=1}^{n}k^2 &= (n+1)^3 - 1^3 - 3\sum_{k=1}^{n}k - n \\
& &= (n+1)^3 - (n+1) - 3 \cdot \frac{n(n+1)}{2}\\
& &= \frac{1}{2}(n+1)\{2(n+1)^2 - 2 - 3n\}\\
& &= \frac{1}{2}(n+1)\{2n^2 + n\}\\
& &= \frac{1}{2}n(n+1)(2n + 1)\\
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&\sum_{k=1}^{n}k^2 &= \frac{n(n + 1)(2n + 1)}{6}\\
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4. 変数 k^3
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&(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1 \\
&\because (k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1\\
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&\begin{array}{rll}
&2^4 &-1^4 &= 4\cdot1^3 &+ 6\cdot1^2 &+ 4\cdot1 &+ 1\\
&3^4 &-2^4 &= 4\cdot2^3 &+ 6\cdot2^2 &+ 4\cdot2 &+ 1\\
&4^4 &-3^4 &= 4\cdot3^3 &+ 6\cdot3^2 &+ 4\cdot3 &+ 1\\
&&:&:\\
+) &(n+1)^4 &-n^4 &= 4\cdot n^3 &+ 6\cdot n^2 &+ 4\cdot n &+ 1\\
&--- &---- &---- &--- &----&\\
&(n+1)^4 &-1^4 &= 4\sum_{k=1}^{n}k^3 &+ 6\sum_{k=1}^{n}k^2 &+ 4\sum_{k=1}^{n}k &+ n\\
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&\begin{array}{rll}
&4\sum_{k=1}^{n}k^3 &= (n+1)^4 - 1 - n - 6\sum_{k=1}^{n}k^2 - 4\sum_{k=1}^{n}k\\
& &= (n+1)^4 - (n+1) - 6 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2}\\
& &= (n+1)\{(n+1)^3 - 1 - n(2n+1) - 2n\}\\
& &= (n+1)\{(n+1)^3 - (2n^2 +3n +1)\}\\
& &= (n+1)\{(n+1)^3 - (2n+1)(n+1)\}\\
& &= (n+1)^2\{(n+1)^2 - (2n+1)\}\\
& &= n^2(n+1)^2\\
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&\sum_{k=1}^{n}k^3 &= \frac{n^2(n + 1))^2}{4} = \{\frac{n(n+1)}{2}\}^2\\
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