1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+...+n(n +1)的值
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1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+...+n(n+1)的值1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+
1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+...+n(n +1)的值
1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+...+n(n +1)的值
1^2+2^2+...+n^2=[n(n+1)(2n+1)]/6,由此求1*2+2*3+...+n(n +1)的值
1*2+2*3+...+n(n +1)
=1²+1+2²+2+.+n²+n
=(1^2+2^2+...+n^2)+(1+2+3+.+n)
=[n(n+1)(2n+1)]/6+n(n+1)/2
=n(n+1)/6 *[2n+1+3]
=n(n+1)(n+2)/3
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