move notebooks to subdir
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notebooks/problem0006.ipynb
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notebooks/problem0006.ipynb
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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "a39ba505",
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"metadata": {},
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"source": [
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"# [Sum Square Difference](https://projecteuler.net/problem=6)\n",
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"\n",
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"In [problem 1](https://projecteuler.net/problem=1), we applied the following formula for [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number):\n",
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"$$1 + 2 + 3 + \\cdots + n = \\frac{n(n+1)}{2}$$\n",
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"We can apply it again here and determine that\n",
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"$$(1 + 2 + 3 + \\cdots + 100)^2 = \\left(\\frac{100(101)}{2}\\right)^2 = 25502500$$\n",
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"\n",
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"A similar formula exists for computing sums of squares, also called the [square pyramidal numbers](https://en.wikipedia.org/wiki/Square_pyramidal_number):\n",
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"$$1^2 + 2^2 + 3^2 + \\cdots + n^2 = \\frac{n(n+1)(2n+1)}{6}$$\n",
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"(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",
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"$$1^2 + 2^2 + 3^2 + \\cdots + 100^2 = \\frac{100(101)(201)}{6} = 338350$$\n",
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"\n",
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"Therefore, the difference is $25502500 - 338350 = 25164150$.\n",
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"\n",
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"## Relevant sequences\n",
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"* Triangular numbers: [A000217](https://oeis.org/A000217)\n",
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"* Square pyramidal numbers: [A000330](https://oeis.org/A000330)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "SageMath 9.5",
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"language": "sage",
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"name": "sagemath"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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