{"id":10604,"date":"2025-10-22T17:58:12","date_gmt":"2025-10-22T08:58:12","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=10604"},"modified":"2026-01-01T10:52:17","modified_gmt":"2026-01-01T01:52:17","slug":"%e6%95%b0%e5%88%97%e3%81%ae%e5%92%8c%e3%81%ae%e5%85%ac%e5%bc%8f%e3%81%a8%e3%81%9d%e3%81%ae%e8%a8%bc%e6%98%8e","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/10604\/","title":{"rendered":"\u6570\u5217\u306e\u548c\u306e\u516c\u5f0f\u3068\u305d\u306e\u8a3c\u660e"},"content":{"rendered":"
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SymPy \u3067\u6570\u5217\u306e\u548c $\\displaystyle \\sum_{k=1}^n k^m \\quad (m = 1, 2, 3, \\dots)$ \u306e\u516c\u5f0f\u304c\u308f\u304b\u308b\u3088\uff0c\u3068\u8aac\u660e\u3057\u305f\u3089\uff0c\u305d\u306e\u8a3c\u660e\u3092\u6c42\u3081\u3089\u308c\u305d\u3046\u306a\u306e\u3067\u30e1\u30e2\u3002<\/p>\n

\u3053\u306e\u516c\u5f0f\u3068\u8a3c\u660e\u306f\u9ad8\u6821\u306e\u6570\u5b66\u3042\u305f\u308a\u3067\u3084\u3063\u3066\u3044\u308b\u3093\u3060\u308d\u3046\u3051\u3069\uff0c\uff08\u306a\u3093\u305b\u662d\u548c\u306e\u6614\u306e\u51fa\u6765\u4e8b\u306a\u306e\u3067\uff09\u79c1\u306b\u306f\u3068\u3093\u3068\u8a18\u61b6\u304c\u306a\u3044\u306a\u3041\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n

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In\u00a0[1]:<\/div>\n
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from<\/span> sympy.abc<\/span> import<\/span> *<\/span>\r\nfrom<\/span> sympy<\/span> import<\/span> *<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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$\\displaystyle\\sum_{k=1}^n k$ = summation(k, (k, 1, n))<\/code><\/h3>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[2]:<\/div>\n
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summation<\/span>(<\/span>k<\/span>,<\/span> (<\/span>k<\/span>,<\/span> 1<\/span>,<\/span> n<\/span>))<\/span>.<\/span>factor<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[2]:<\/div>\n
$\\displaystyle \\frac{n \\left(n + 1\\right)}{2}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u8a3c\u660e<\/h4>\n
\\begin{eqnarray}
\n\\sum_{k=1}^n k &=& \\frac{1}{2} \\biggl\\{\\ 1 + 2 + \\cdots + (n-1) + n \\\\
\n&& \\quad\u00a0 \\\u00a0 + n + (n-1) + \\cdots + 2 + 1 \\biggr\\} \\\\
\n&=& \\frac{1}{2} \\sum_{k=1}^n \\left(n+1\\right)\\\\
\n&=& \\frac{1}{2} n (n+1)
\n\\end{eqnarray}<\/p>\n
\u3067\u3082\u3053\u308c\u3060\u3068\uff0c\u6b21\u306e $\\displaystyle \\sum_{k=1}^n k^2$ \u3084 $\\displaystyle \\sum_{k=1}^n k^3$ \u306e\u3068\u304d\u306b\u306f\u5fdc\u7528\u3067\u304d\u306a\u3044\u306e\u3067\uff0c\u5225\u89e3\uff1a<\/p>\n
\\begin{align}
\n& &(k+1)^2 -k^2 &= 2 k + 1\\\\ \\ \\\\
\n&k=1: & 2^2 -1^2 &= 2\\cdot 1 + 1 \\\\
\n&k=2: & 3^2 -2^2 &= 2\\cdot 2 + 1 \\\\
\n&\\quad \\vdots & &\\ \\ \\vdots\\\\
\n&k=n-1: & n^2 -(n-1)^2 &= 2\\cdot(n-1) + 1\\\\
\n&k=n: &(n+1)^2 -n^2 &= 2\\cdot n + 1
\n\\end{align}
\n$k=1$ \u304b\u3089 $k=n$ \u307e\u3067\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u8db3\u3057\u5408\u308f\u305b\u308b\u3068\uff0c\u5de6\u8fba\u306f\u6700\u5f8c\u306e $(n+1)^2$ \u3068\u6700\u521d\u306e $-1^2$ \u306e\u9805\u4ee5\u5916\u306f\u5168\u3066\u30ad\u30e3\u30f3\u30bb\u30eb\u3055\u308c\u308b\u304b\u3089…
\n\\begin{align}
\n(n+1)^2 -1 &= 2 \\sum_{k=1}^n k + n \\\\
\n\\therefore \\ \\ n^2 + 2 n &= 2 \\sum_{k=1}^n k + n \\\\
\n\\therefore\\ \\ \\sum_{k=1}^n k &= \\frac{n(n+1)}{2}
\n\\end{align}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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$\\displaystyle\\sum_{k=1}^n k^2$ = summation(k**2, (k, 1, n))<\/code><\/h3>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[3]:<\/div>\n
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summation<\/span>(<\/span>k<\/span>**<\/span>2<\/span>,<\/span> (<\/span>k<\/span>,<\/span> 1<\/span>,<\/span> n<\/span>))<\/span>.<\/span>factor<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[3]:<\/div>\n
$\\displaystyle \\frac{n \\left(n + 1\\right) \\left(2 n + 1\\right)}{6}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u8a3c\u660e<\/h4>\n
\\begin{align}
\n& &(k+1)^3 -k^3 &= 3 k^2 + 3 k + 1\\\\ \\ \\\\
\n&k=1: & 2^3 -1^3 &= 3\\cdot 1^2 + 3\\cdot 1 + 1 \\\\
\n&k=2: & 3^3 -2^3 &= 3\\cdot 2^2 + 3\\cdot 2 + 1 \\\\
\n&\\quad\\vdots & &\\ \\ \\vdots\\\\
\n&k=n-1: & n^3 -(n-1)^3 &= 3\\cdot(n-1)^2 + 3\\cdot (n-1) + 1\\\\
\n&k=n: &(n+1)^3 -n^3 &= 3\\cdot n^2 + 3\\cdot n + 1
\n\\end{align}
\n$k=1$ \u304b\u3089 $k=n$ \u307e\u3067\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u8db3\u3057\u5408\u308f\u305b\u308b\u3068\uff0c\u5de6\u8fba\u306f\u6700\u5f8c\u306e $(n+1)^3$ \u3068\u6700\u521d\u306e $-1^3$ \u306e\u9805\u4ee5\u5916\u306f\u5168\u3066\u30ad\u30e3\u30f3\u30bb\u30eb\u3055\u308c\u308b\u304b\u3089…
\n\\begin{align}
\n(n+1)^3 -1 &= 3 \\sum_{k=1}^n k^2 + 3 \\sum_{k=1}^n k + n \\\\
\n\\therefore \\ \\ n^3 + 3 n^2 + 3 n &= 3 \\sum_{k=1}^n k^2 + 3\\cdot\\frac{n(n+1)}{2} + n \\\\
\n\\therefore\\ \\ \\sum_{k=1}^n k^2 &= \\frac{1}{3}\\left\\{n^3 + 3 n^2 + 3 n -3\\frac{n(n+1)}{2} -n\\right\\}\\\\
\n&= \\frac{n(n+1)(2 n+1)}{6}
\n\\end{align}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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$\\displaystyle\\sum_{k=1}^n k^3$ = summation(k**3, (k, 1, n))<\/code><\/h3>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[4]:<\/div>\n
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summation<\/span>(<\/span>k<\/span>**<\/span>3<\/span>,<\/span> (<\/span>k<\/span>,<\/span> 1<\/span>,<\/span> n<\/span>))<\/span>.<\/span>factor<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[4]:<\/div>\n
$\\displaystyle \\frac{n^{2} \\left(n + 1\\right)^{2}}{4}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u8a3c\u660e<\/h4>\n
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\u8a3c\u660e\u306f… \uff08$(k+1)^4$ \u306e\u5c55\u958b\u306f SymPy \u306b\u3084\u3063\u3066\u3082\u3089\u3046\u3053\u3068\u306b\u3057\u3066\uff09<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[5]:<\/div>\n
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Eq<\/span>((<\/span>k<\/span>+<\/span>1<\/span>)<\/span>**<\/span>4<\/span> -<\/span> k<\/span>**<\/span>4<\/span>,<\/span> expand<\/span>((<\/span>k<\/span>+<\/span>1<\/span>)<\/span>**<\/span>4<\/span> -<\/span> k<\/span>**<\/span>4<\/span>))<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[5]:<\/div>\n
$\\displaystyle -k^{4} + \\left(k + 1\\right)^{4} = 4 k^{3} + 6 k^{2} + 4 k + 1$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u3067\u3059\u304b\u3089\uff0c$k=1$ \u304b\u3089 $k=n$ \u307e\u3067\u306e
\n$$(k+1)^4 -k^4 = 4 k^3 + 6 k^2 + 4 k + 1$$
\n\u306e\u4e21\u8fba\u3092\u8db3\u3057\u5408\u308f\u305b\u308c\u3070\u8a3c\u660e\u3067\u304d\u307e\u3059\u3002<\/p>\n
$k=1$ \u304b\u3089 $k=n$ \u307e\u3067\u306e\u5f0f\u306e\u4e21\u8fba\u3092\u8db3\u3057\u5408\u308f\u305b\u308b\u3068\uff0c\u5de6\u8fba\u306f\u6700\u5f8c\u306e $(n+1)^4$ \u3068\u6700\u521d\u306e $-1^4$ \u306e\u9805\u4ee5\u5916\u306f\u5168\u3066\u30ad\u30e3\u30f3\u30bb\u30eb\u3055\u308c\u308b\u304b\u3089…
\n\\begin{align}
\n(n+1)^4 -1 &= 4 \\sum_{k=1}^n k^3 + 6 \\sum_{k=1}^n k^2 + 4 \\sum_{k=1}^n k + n \\\\
\n\\therefore \\ \\ n^4 + 4 n^3 + 6 n^2 + 4 n &= 4 \\sum_{k=1}^n k^3 + 6\\cdot\\frac{n(n+1)(2 n+1)}{6} + 4\\cdot\\frac{n(n+1)}{2} + n \\\\
\n\\therefore\\ \\ \\sum_{k=1}^n k^3 &= \\frac{1}{4}\\left\\{n^4 + 4 n^3 + 6 n^2 + 4 n -n(n+1)(2 n+1) -2n(n+1) -n\\right\\}\\\\
\n&= \\frac{n^4 + 2 n^3 + n^2}{4} \\\\
\n&= \\frac{n^2 (n+1)^2}{4}
\n\\end{align}<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
SymPy \u3067\u6570\u5217\u306e\u548c $\\displaystyle \\sum_{k=1}^n k^m \\quad (m = 1, 2, 3, \\dots)$ \u306e\u516c\u5f0f\u304c\u308f\u304b\u308b\u3088\uff0c\u3068\u8aac\u660e\u3057\u305f\u3089\uff0c\u305d\u306e\u8a3c\u660e\u3092\u6c42\u3081\u3089\u308c\u305d\u3046\u306a\u306e\u3067\u30e1\u30e2\u3002<\/p>\n
\u3053\u306e\u516c\u5f0f\u3068\u8a3c\u660e\u306f\u9ad8\u6821\u306e\u6570\u5b66\u3042\u305f\u308a\u3067\u3084\u3063\u3066\u3044\u308b\u3093\u3060\u308d\u3046\u3051\u3069\uff0c\uff08\u306a\u3093\u305b\u662d\u548c\u306e\u6614\u306e\u51fa\u6765\u4e8b\u306a\u306e\u3067\uff09\u79c1\u306b\u306f\u3068\u3093\u3068\u8a18\u61b6\u304c\u306a\u3044\u306a\u3041\u3002<\/p>
\u7d9a\u304d\u3092\u8aad\u3080<\/a><\/p>\n","protected":false},"author":33,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[12],"tags":[],"class_list":["post-10604","post","type-post","status-publish","format-standard","hentry","category-sympy","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t