Question

∑(i=1到n)(1-i/100)*(i/100)怎么算？

Answer 1

∑(i=1到n)(1-i/100)*(i/100)
=∑(i=1到n)[i/100-i^2/10000]
=n(1+n)/200-n(n+1)(2n+1)/60000
=-n(n+1)(2n-299)/60000.

Answer 2

∑(i=1->n)(1- i/100)(i/100)
=(1/10000)∑(i=1->n) (100-i)i
=(1/10000) ∑(i=1->n) (-i^2+100i)
=(1/10000) ∑(i=1->n) (-i(i+1)+101i)
=(1/10000) ∑(i=1->n) { -(1/3)[i(i+1)(i+2)-(i-1)i(i+1)]+ (101/2)[ i(i+1)-(i-1)i] }
=(1/10000) [ -(1/3)n(n+1)(n+2)+ (101/2)n(n+1) ]
=(1/10000) [ -(1/6)n(n+1) [ 2(n+2)- 303 ]
=-(1/60000) n(n+1) (2n-299)