{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We know that the formula for the [sum of squares](https://en.wikipedia.org/wiki/Square_pyramidal_number) is $\\sum_{k=1}^{n} k^2 = \\frac{2n^3 + 3n^2 + n}{6}$ and that $\\sum_{k=1}^{n} k = \\frac{n*(n+1)}{2}$ (see [Arithmetic progression](https://en.wikipedia.org/wiki/Arithmetic_progression)). So we simply need to do $\\sum_{k=1}^{n}k^2 - (\\sum_{k=1}^{n}k)^2 = \\frac{3n^4 + 2n^3 - 3n^2 - 2n}{12}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "25164150"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def equation(n):\n",
    "    return (3*n**4 + 2*n**3 - 3*n**2 - 2*n) // 12\n",
    "equation(100)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.7.4"
  },
  "title": "Sum square difference"
 },
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 "nbformat_minor": 2
}