1*1+2*3+3*5+4*7+...+49*97+50*99等于多少

Question

1*1+2*3+3*5+4*7+...+49*97+50*99等于多少

Answer 1

解：
考察一般项：k(2k-1)=2k²-k
1×2+2×3+3×5+...+49×97+50×99
=2×(1²+2²+3²+...+50²)-(1+2+3+...+50)
=2×50×(50+1)×(2×50+1)/6 -50×(50+1)/2
=84575

一般的：
1×2+2×3+...+n(2n-1)
=2×(1²+2²+...+n²)-(1+2+...+n)
=2n(n+1)(2n+1)/6 -n(n+1)/2
=[n(n+1)/6][2(2n+1)-3]
=[n(n+1)/6](4n-1)
=n(n+1)(4n-1)/6本回答被网友采纳

Answer 2

原式=∑(n=1~50) n*(2n-1)
=∑(n=1~50) (2n^2-n)
=2*∑(n=1~50) n^2-∑(n=1~50) n
=2*[50*(50+1)*(2*50+1)/6]-(50+1)*50/2
=84575