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Worked solution

Correct summation bounds by adding or removing terms

Add or subtract the missing terms to make each equation valid: (a) ∑r=2nr2=∑r=1nr2−□\sum_{r=2}^{n}r^2=\sum_{r=1}^{n}r^2-\square; (b) ∑r=5101/(r+1)=∑r=2101/(r+1)−□\sum_{r=5}^{10}1/(r+1)=\sum_{r=2}^{10}1/(r+1)-\square; (c) ∑n=1N2/n=∑n=3N2/n+□\sum_{n=1}^{N}2/n=\sum_{n=3}^{N}2/n+\square.

CJC · Sequences and Series · Example 16

(a) Remove the first square

Answer

121^2.

Remove only the extra index

∑r=2nr2=∑r=1nr2−12\sum_{r=2}^{n}r^2=\sum_{r=1}^{n}r^2-1^2

The right-hand sum includes r=1, while the required sum begins at r=2. Here n is an integer at least 2.

How did this part go?

(b) Remove the first three reciprocal terms

Answer

∑r=241r+1\sum_{r=2}^{4}\frac1{r+1}.

Stop the removed prefix just before the new lower bound

∑r=5101r+1=∑r=2101r+1−∑r=241r+1\begin{aligned}&\sum_{r=5}^{10}\frac1{r+1}\\ & =\sum_{r=2}^{10}\frac1{r+1}-\sum_{r=2}^{4}\frac1{r+1}\end{aligned}

Remove indices 2,3,4, not the first required index 5.

How did this part go?

(c) Restore the first two terms

Answer

∑n=122n\sum_{n=1}^{2}\frac2n.

Add the indices excluded from the tail

∑n=1N2n=∑n=3N2n+∑n=122n\begin{aligned}&\sum_{n=1}^{N}\frac2n\\ & =\sum_{n=3}^{N}\frac2n+\sum_{n=1}^{2}\frac2n\end{aligned}

The source uses n as the counter and N as the upper limit. Here N is an integer at least 3.

How did this part go?