How do you find the partial sum of #Sigma (4.5+0.025j)# from j=1 to 200?

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Answer 1 · 2018-06-24T06:47:35

# 1402.5#.

We know that, #sum_(j=1)^nkj=k/2n(n+1), &, sum_(j=1)^nk=kn#.

#:. sum_(j=1)^200(4.5+0.025j)#,

#=sum_(j=1)^200 4.5+sum_(j=1)^200 0.025j#,

#=(4.5)(200)+0.025sum_(j=1)^200 j#,

#=900+0.025{1/2*200*(200+1)}#,

#=900+502.5#,

#=1402.5#.