![\sum^{25}_{16}i^2=\sum^{25}_{0}i^2-\sum^{15}_{0}i^2\\\\\text{Use the summation formula:}\ \dfrac{n}{6}(n+1)(2n+1)\\\\\sum^{25}_{0}i^2 = \dfrac{25}{6}(25+1)[2(25)+1]\\\\\\.\qquad=\dfrac{25(26)(51)}{6}\\\\.\qquad=5525\\\\\\\sum^{15}_{0}i^2 = \dfrac{15}{6}(15+1)[2(15)+1]\\\\\\.\qquad=\dfrac{15(16)(31)}{6}\\\\.\qquad=1240\\\\\\\sum^{25}_{0}i^2-\sum^{15}_{0}i^2\\\\=5525-1240\\\\=4285](/tpl/images/0010/4531/6ec38.png)
Find the sum of the g. p,16^2+17^2+18^2++25^2
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Mathematics, 24.06.2019 04:00 Domi0129
Find the sum of the g. p,16^2+17^2+18^2++25^2
Answers: 2
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