Porisma

Worked solution

Shift both bounds before applying a prefix formula

Given ∑r=2n−1[2r+4(1/(r+1)−1/r)]=n2−n+4/n−4\sum_{r=2}^{n-1}[2r+4(1/(r+1)-1/r)]=n^2-n+4/n-4, show

∑r=92n−2[2r+2+4(1r+2−1r+1)]=4n2−2n+2n−4525.\sum_{r=9}^{2n-2}\left[2r+2+4\left(\frac1{r+2}-\frac1{r+1}\right)\right]=4n^2-2n+\frac2n-\frac{452}5.
marks

[2013/SRJC/Promo/10(modified)]

CJC · Review problems · Question 7

Answer

4n2−2n+2/n−452/54n^2-2n+2/n-452/5.

Change index to match the supplied summand

Put k=r+1k=r+1. The summand becomes 2k+4(1/(k+1)−1/k)2k+4(1/(k+1)-1/k), and the bounds change from r=9,2n−2r=9,2n-2 to k=10,2n−1k=10,2n-1.

How did this part go?

Subtract the prefix with upper index 9

Let F(m)=m2−m+4/m−4F(m)=m^2-m+4/m-4, the sum from 22 to m−1m-1. The required sum is F(2n)−F(10)=[4n2−2n+2/n−4]−[100−10+4/10−4]=4n2−2n+2/n−452/5F(2n)-F(10)=[4n^2-2n+2/n-4]-[100-10+4/10-4]=4n^2-2n+2/n-452/5. The ordinary nonempty range uses integer n≥6n\ge6.

How did this part go?

Common mistakes

The supplied prefix ends at m−1, so an upper limit of 2n−1 requires m=2n.

Read the hint