已知1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),求2²+4²+…+50²
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![已知1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),求2²+4²+…+50²](/uploads/image/z/2689011-27-1.jpg?t=%E5%B7%B2%E7%9F%A51%26%23178%3B%3B%2B2%26%23178%3B%3B%2B3%26%23178%3B%3B%2B%E2%80%A6%2Bn%26%23178%3B%3B%3D1%2F6n%28n%2B1%29%282n%2B1%29%2C%E6%B1%822%26%23178%3B%2B4%26%23178%3B%2B%E2%80%A6%2B50%26%23178%3B)
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已知1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),求2²+4²+…+50²
已知1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),求2²+4²+…+50²
已知1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),求2²+4²+…+50²
2²+4²+…+50²=4(1²+2²+3²+…+25²)=2/3*25*(25+1)(50+1)=226100
或
(2²+4²+…+50² )可看成由(1²;+2²;+3²;+…+25²)整个乘以4得到.因为1²;+2²;+3²;+…+n²;=1/6n(n+1)(2n+1),所以将n=25代入1/6n(n+1)(2n+1),算出得数后,再乘以四就行了