Porisma

Change bounds and combine sums

Count terms, remove prefixes and match a given summation identity.

Apply and prove summation laws

For finite integer bounds m≤nm\le n and constant cc, a constant factor can be taken outside a sum:

∑r=mncur=cum+cum+1+⋯+cun=c(um+um+1+⋯+un−1+un)=c∑r=mnur.\begin{aligned} \sum_{r=m}^{n}cu_r &=cu_m+cu_{m+1}+\cdots+cu_n\\ &=c(u_m+u_{m+1}+\cdots+u_{n-1}+u_n)\\ &=c\sum_{r=m}^{n}u_r. \end{aligned}

The source’s examples keep their original bounds:

∑r=1102r=2∑r=110r,∑r=121003r2=3∑r=12100r2.\sum_{r=1}^{10}2r=2\sum_{r=1}^{10}r,\qquad \sum_{r=12}^{100}3r^2=3\sum_{r=12}^{100}r^2.

Sums and differences can also be separated over the same bounds. The source proves addition by expanding and regrouping:

∑r=mn(ur+vr)=(um+vm)+(um+1+vm+1)+⋯+(un−1+vn−1)+(un+vn)=(um+um+1+⋯+un−1+un)+(vm+vm+1+⋯+vn−1+vn)=∑r=mnur+∑r=mnvr.\begin{aligned} \sum_{r=m}^{n}(u_r+v_r) &=(u_m+v_m)+(u_{m+1}+v_{m+1})+\cdots\\ &\quad+(u_{n-1}+v_{n-1})+(u_n+v_n)\\ &=(u_m+u_{m+1}+\cdots+u_{n-1}+u_n)\\ &\quad+(v_m+v_{m+1}+\cdots+v_{n-1}+v_n)\\ &=\sum_{r=m}^{n}u_r+\sum_{r=m}^{n}v_r. \end{aligned}

For subtraction, regroup the negative terms with their signs intact:

∑r=mn(ur−vr)=(um−vm)+(um+1−vm+1)+⋯+(un−1−vn−1)+(un−vn)=(um+um+1+⋯+un−1+un)−(vm+vm+1+⋯+vn−1+vn)=∑r=mnur−∑r=mnvr.\begin{aligned} \sum_{r=m}^{n}(u_r-v_r) &=(u_m-v_m)+(u_{m+1}-v_{m+1})+\cdots\\ &\quad+(u_{n-1}-v_{n-1})+(u_n-v_n)\\ &=(u_m+u_{m+1}+\cdots+u_{n-1}+u_n)\\ &\quad-(v_m+v_{m+1}+\cdots+v_{n-1}+v_n)\\ &=\sum_{r=m}^{n}u_r-\sum_{r=m}^{n}v_r. \end{aligned}

The two printed examples are

∑r=13(r+er)=∑r=13r+∑r=13er,\sum_{r=1}^{3}(r+e^r)=\sum_{r=1}^{3}r+\sum_{r=1}^{3}e^r, ∑r=1288(r−r2)=∑r=1288r−∑r=1288r2.\sum_{r=12}^{88}(r-r^2)=\sum_{r=12}^{88}r-\sum_{r=12}^{88}r^2.

These rules do not say that a sum of products equals the product of two sums.

Remove a prefix and restore missing terms

To isolate a tail when k<m≤nk<m\le n, subtract the excluded prefix:

∑r=mnur=∑r=knur−∑r=km−1ur.\sum_{r=m}^{n}u_r=\sum_{r=k}^{n}u_r-\sum_{r=k}^{m-1}u_r.

The source proves this by adding and subtracting the same prefix. Write its two groups as

A=uk+uk+1+⋯+um−2+um−1,B=um+um+1+⋯+un−1+un.\begin{aligned} A&=u_k+u_{k+1}+\cdots+u_{m-2}+u_{m-1},\\ B&=u_m+u_{m+1}+\cdots+u_{n-1}+u_n. \end{aligned}

Then its cancellation argument is

∑r=mnur=B=A+B−A=(uk+uk+1+⋯+um−2+um−1+um+um+1+⋯+un−1+un)−A=∑r=knur−∑r=km−1ur.\begin{aligned} \sum_{r=m}^{n}u_r&=B=A+B-A\\ &=(u_k+u_{k+1}+\cdots+u_{m-2}+u_{m-1}\\ &\quad+u_m+u_{m+1}+\cdots+u_{n-1}+u_n)-A\\ &=\sum_{r=k}^{n}u_r-\sum_{r=k}^{m-1}u_r. \end{aligned}

The letters A and B organise the same printed expansion into shorter lines. As with any ellipsis, include each actual term once when the ranges are short. The source notes that this is useful when a formula for the prefix from 1 is given but the required sum starts at k+1k+1: subtract the first k terms.

Both printed rule examples retain their exact excluded prefixes:

∑r=410r=∑r=110r−∑r=13r,\sum_{r=4}^{10}r=\sum_{r=1}^{10}r-\sum_{r=1}^{3}r, ∑r=101001r(r−1)=∑r=21001r(r−1)−∑r=291r(r−1).\sum_{r=10}^{100}\frac1{r(r-1)} =\sum_{r=2}^{100}\frac1{r(r-1)}-\sum_{r=2}^{9}\frac1{r(r-1)}.

Example 16 asks you to supply missing terms: remove 121^2 to move a square sum’s lower bound from 1 to 2; remove ∑r=241/(r+1)\sum_{r=2}^{4}1/(r+1) to move a reciprocal sum from 2 to 5; add ∑n=122/n\sum_{n=1}^{2}2/n to restore the first two terms of a sum starting at 3. Adding or subtracting terms is another way to obtain equivalent sigma expressions, alongside changing the counter.

Match a shifted sum to a given identity

Example 17 gives

∑n=1N1n(n+1)(n+2)=14−12(N+1)+12(N+2)\sum_{n=1}^{N}\frac1{n(n+1)(n+2)} =\frac14-\frac1{2(N+1)}+\frac1{2(N+2)}

and asks for ∑n=3N1/[n(n−1)(n−2)]\sum_{n=3}^{N}1/[n(n-1)(n-2)], with integer N≥3N\ge3. Its two methods make the same change visible in different ways.

List the terms: the first terms have denominators 3⋅2⋅13\cdot2\cdot1, 4⋅3⋅24\cdot3\cdot2, 5⋅4⋅35\cdot4\cdot3. Reorder each product to match 1⋅2⋅31\cdot2\cdot3, 2⋅3⋅42\cdot3\cdot4, 3⋅4⋅53\cdot4\cdot5, and continue through (N−2)(N−1)N(N-2)(N-1)N. There are N−2N-2 terms, so apply the supplied identity with upper bound N−2, not N.

Change the counter: let r=n−2r=n-2, giving n−1=r+1n-1=r+1 and n=r+2n=r+2. The illustrated source guidance uses a new letter to prevent confusion and aims to match the given formula with lower index 1. Transform both endpoints: n=3n=3 gives r=1r=1, while n=Nn=N gives r=N−2r=N-2. Thus

∑n=3N1n(n−1)(n−2)=∑r=1N−21(r+2)(r+1)r=∑r=1N−21r(r+1)(r+2)=14−12[(N−2)+1]+12[(N−2)+2]=14−12(N−1)+12N.\begin{aligned} \sum_{n=3}^{N}\frac1{n(n-1)(n-2)} &=\sum_{r=1}^{N-2}\frac1{(r+2)(r+1)r}\\ &=\sum_{r=1}^{N-2}\frac1{r(r+1)(r+2)}\\ &=\frac14-\frac1{2[(N-2)+1]}+\frac1{2[(N-2)+2]}\\ &=\frac14-\frac1{2(N-1)}+\frac1{2N}. \end{aligned}

The full worked exercise retains both printed methods, including the final two list terms and the Teacher’s continuation of the endpoint substitution.

Sum a constant by counting its terms

Multiplication is repeated addition. The source’s opening example has eight copies of 5:

5+5+5+5+5+5+5+5=∑r=185=8×5.5+5+5+5+5+5+5+5=\sum_{r=1}^{8}5=8\times5.

For finite integer bounds m≤nm\le n and constant cc,

∑r=mnc=c+c+⋯+c⏟n−m+1 terms=(n−m+1)c.\sum_{r=m}^{n}c=\underbrace{c+c+\cdots+c}_{n-m+1\text{ terms}}=(n-m+1)c.

The factor counts how many times c is repeated. Example 18 includes a negative starting index and a negative summand: ∑r=−275=[7−(−2)+1]5=10(5)=50\sum_{r=-2}^{7}5=[7-(-2)+1]5=10(5)=50, while ∑r=310(−2)=(10−3+1)(−2)=8(−2)=−16\sum_{r=3}^{10}(-2)=(10-3+1)(-2)=8(-2)=-16.

Try these examples

Open an example in chapter practice. Try it first, then read its solution. Use “Learn this step” to return to this unit.

Apply this idea

An integer upper limit in an inequality →Evaluate the arithmetic sum, solve the inequality and enforce its integer indexing restriction.