Calculating sums indirectly

Section 5.4 Calculating sums indirectly

In the previous section, we were able to calculate the sums directly because we knew how to calculate each term of the relevant sequences. This isn't always the case. Sometimes we will need to be able to manipulate sums where the individual terms of the sequence are unknown. In particular, one of the tools we will learn about in this section will be of paramount importance when we learn to write proofs by induction later.

Theorem 5.4.1. Linear property of summation.

Let \((a_n)\) and \((b_n)\) be sequences and \(c\) be a constant value. Then,

\begin{equation*} \sum_{n=k}^m (a_n + b_n) = \sum_{n=k}^m a_n + \sum_{n=k}^m b_n \end{equation*}

and

\begin{equation*} \sum_{n=k}^m ca_n = c\sum_{n=k}^m a_n \end{equation*}

The above theorem is just a generalization of the associative property of addition—\((x+y)+z = x+(y+z)\) —and the distributive property—\(c(x+y) = cx+cy\text{.}\)

Example 5.4.2.

Suppose we know that

\begin{equation*} \sum_{n=2}^{38} a_n = 14 \end{equation*}

and

\begin{equation*} \sum_{n=2}^{38} b_n = 19 \end{equation*}

Calculate

\begin{equation*} \sum_{n=2}^{38} (2a_n-b_n) \end{equation*}

Using the linear property,

\begin{align*} \sum_{n=2}^{38} (2a_n-b_n) \amp = 2\sum_{n=2}^{38} a_n - \sum_{n=2}^{38} b_n\\ \amp = 2(14) - 19\\ \amp = 28 - 9\\ \amp = 9 \end{align*}

We can calculate this sum without having any idea what the terms of the sequences \((a_n)\) and \((b_n)\) are!

Notice that both the sums must start and end on the same index. If not, you can manipulate the sums to make the linear property useable, but it reqires knowing some terms of the sequence.

Theorem 5.4.3. Recursive property of summation.

Let \((a_n)\) be a sequence. Then

\begin{align*} \sum_{n=k}^m a_n \amp = \left( \sum_{n=k}^{m-1} a_n \right) + a_m\\ \amp = a_k + \left( \sum_{n=k+1}^{m} a_n \right) \end{align*}

In other words, we may “peel” the first or last term off of a sum (as many times as we like).

Proof.

This theorem is a consequence of the associative property of addition. By definition,

\begin{equation*} \sum_{n=k}^m a_n = a_k + a_{k+1} + a_{k+2} + \cdots + a_{m-1} + a_m \end{equation*}

We may group all but the last term together:

\begin{equation*} \sum_{n=k}^m a_n = \left( a_k + a_{k+1} + a_{k+2} + \cdots + a_{m-1} \right) + a_m \end{equation*}

We may rewrite the grouped terms as a single sum.

\begin{equation*} \sum_{n=k}^m a_n = \left( \sum_{n=k}^{m-1} a_n \right) + a_m \end{equation*}

The proof that

\begin{equation*} \sum_{n=k}^m a_n = a_k + \left( \sum_{n=k+1}^{m} a_n \right) \end{equation*}

follows in a similar fashion.

Example 5.4.4.

Suppose it is known that \(b_{107} = 72\) and that

\begin{equation*} \sum_{k=1}^{106} b_k = 291 \end{equation*}

Then,

\begin{align*} \sum_{k=1}^{107} b_k \amp = \left( \sum_{k=1}^{106} b_k \right) + b_{107}\\ \amp = 291 + 72\\ \amp = 363 \end{align*}

Example 5.4.5.

Suppose we know \(a_3=12\text{,}\) \(a_4=2\text{,}\) \(a_5 = 7\text{,}\) \(a_6=-3\text{,}\) \(a_{13}=19\text{,}\) \(a_{14}=-10\text{,}\) and \(a_{15}=6\text{,}\) and furthermore that

\begin{equation*} \sum_{k=7}^{15} a_k = 109 \end{equation*}

What is the value of

\begin{equation*} \sum_{k=4}^{13} a_k? \end{equation*}

Our goal is to write the unknown sum in terms of the known sum. We can start by writing the unknown sum out the long way.

\begin{equation*} \sum_{k=4}^{13} a_k = a_4 + a_5 + a_6 + a_7 + a_8 + \cdots + a_{12} + a_{13} \end{equation*}

For reference, our known sum is

\begin{equation*} \sum_{k=7}^{15} a_k = a_7 + a_8 + \cdots + a_{12} + a_{13} + a_{14} + a_{15} \end{equation*}

We have many of these terms in our unknown sum, but not all. The great news is that we can always add whatever value we want to an expression, at the cost of subtracting that same value. So, back to the unknown sum:

\begin{align*} \sum_{k=4}^{13} a_k \amp = a_4 + a_5 + a_6 + a_7 + a_8 + \cdots + a_{12} + a_{13} + a_{14} + a_{15} - a_{14} - a_{15}\\ \amp = a_4 + a_5 + a_6 + \big( a_7 + a_8 + \cdots + a_{12} + a_{13} + a_{14} + a_{15} \big) - a_{14} - a_{15}\\ \amp = a_4 + a_5 + a_6 + \big( \sum_{k=7}^{15} a_k \big) - a_{14} - a_{15}\\ \amp = 2 + 7 + (-3) + 109 - (-10) - 6\\ \amp = 119 \end{align*}

The moral here is that with careful application of the recursive property, we can shift the indices of our sum to whatever we like, as long as we know enough terms of the sequence.

Example 5.4.6.

Given that

\begin{equation*} \sum_{i=1}^{13} a_i = 90, \end{equation*}

\(a_1=7\text{,}\) \(b_{13}=18\text{,}\) and

\begin{equation*} \sum_{i=2}^{12} b_i = 13, \end{equation*}

calculate

\begin{equation*} \sum_{i=2}^{13} (4a_i + 3b_i) \end{equation*}

Start by using the linear property of the sum. Then, we will need to evaluate the two individual sums recursively.

\begin{align*} \sum_{i=2}^{13} (4a_i + 3b_i) \amp = 4 \sum_{i=2}^{13} a_i + 3 \sum_{i=2}^{13}b_i\\ \amp = 4\big(\sum_{i=1}^{13} a_i - a_1\big) + 3\big(\sum_{i=2}^{12} b_i + b_{13}\big)\\ \amp = 4(90-7) + 3(13+18)\\ \amp = 332 + 93\\ \amp = 425 \end{align*}