Author: fireheartjerry
Compute the last 3 digits of the sum 1^2+2^2+3^2+...+69^2.
There’s a formula for this as well!
", "editorial_md": "The formula for the sum of squares of the first $n$ natural numbers is:\n$$\n\\displaystyle\\sum_{k=1}^{n} k^2 = \\frac{n(n+1)(2n+1)}{6}\n$$\n\nGiven that $n=69$, we want to find the sum of squares from 1 to 69 and then determine the last three digits of that sum. This means we need to compute:\n$$\n\\displaystyle\\sum_{k=1}^{n} k^2 = \\frac{69\\cdot(69+1)\\cdot(2\\cdot69+1)}{6}\n$$\nWhich nets us the following:\n$$\n\\displaystyle\\sum_{k=1}^{69} k^2 = \\frac{69\\cdot70\\cdot139}{6}\n$$\nSolving further gives a result of $111795$. From here, because the question is looking for the last 3 digits of this number, the answer is simply $795$.\n\n- botman", "editorial_html": "The formula for the sum of squares of the first n natural numbers is:
\\displaystyle\\sum_{k=1}^{n} k^2 = \\frac{n(n+1)(2n+1)}{6}
Given that n=69, we want to find the sum of squares from 1 to 69 and then determine the last three digits of that sum. This means we need to compute:
\\displaystyle\\sum_{k=1}^{n} k^2 = \\frac{69\\cdot(69+1)\\cdot(2\\cdot69+1)}{6}
Which nets us the following:
\\displaystyle\\sum_{k=1}^{69} k^2 = \\frac{69\\cdot70\\cdot139}{6}
Solving further gives a result of 111795. From here, because the question is looking for the last 3 digits of this number, the answer is simply 795.