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k^2
= k(k+1) -k
= (1/3)[ k(k+1)(k+2) -(k-1)k(k+1)] -(1/2)[k(k+1)-(k-1)k]
∑(k:1->n) k^2
=∑(k:1->n) { (1/3)[ k(k+1)(k+2) -(k-1)k(k+1)] -(1/2)[k(k+1)-(k-1)k] }
=(1/3)n(n+1)(n+2) -(1/2)n(n+1)
=(1/6)n(n+1)( 2n+4 -3)
=(1/6)n(n+1)( 2n+1)
= k(k+1) -k
= (1/3)[ k(k+1)(k+2) -(k-1)k(k+1)] -(1/2)[k(k+1)-(k-1)k]
∑(k:1->n) k^2
=∑(k:1->n) { (1/3)[ k(k+1)(k+2) -(k-1)k(k+1)] -(1/2)[k(k+1)-(k-1)k] }
=(1/3)n(n+1)(n+2) -(1/2)n(n+1)
=(1/6)n(n+1)( 2n+4 -3)
=(1/6)n(n+1)( 2n+1)
追问
都说了必须要用二项式定理做鸭🙃🙃
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