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    "# [Sum Square Difference](https://projecteuler.net/problem=6)\n",
    "\n",
    "In [problem 1](https://projecteuler.net/problem=1), we applied the following formula for [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number):\n",
    "$$1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2}$$\n",
    "We can apply it again here and determine that\n",
    "$$(1 + 2 + 3 + \\cdots + 100)^2 = \\left(\\frac{100(101)}{2}\\right)^2 = 25502500$$\n",
    "\n",
    "A similar formula exists for computing sums of squares, also called the [square pyramidal numbers](https://en.wikipedia.org/wiki/Square_pyramidal_number):\n",
    "$$1^2 + 2^2 + 3^2 + \\cdots + n^2 = \\frac{n(n+1)(2n+1)}{6}$$\n",
    "(In fact, [Faulhaber's formula](https://en.wikipedia.org/wiki/Faulhaber%27s_formula) gives a formula for the sum of $k$th powers, but we obviously only need the cases $k=1$ and $k=2$ for this problem.) Consequently,\n",
    "$$1^2 + 2^2 + 3^2 + \\cdots + 100^2 = \\frac{100(101)(201)}{6} = 338350$$\n",
    "\n",
    "Therefore, the difference is $25502500 - 338350 = 25164150$.\n",
    "\n",
    "## Relevant sequences\n",
    "* Triangular numbers: [A000217](https://oeis.org/A000217)\n",
    "* Square pyramidal numbers: [A000330](https://oeis.org/A000330)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
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