1.1 Sigma notation, finite & Infinite series  (Page 3/4)

Finite squared series

When we sum a finite number of terms in a quadratic sequence, we get a finite quadratic series . The general form of a quadratic series is quite complicated, so we will only look at the simple case when $D=2$ and $d=(a_2-a_1)=3$ , where $D$ is the common second difference and $d$ is the finite difference. This is the sequence of squares of the integers:

If we wish to sum this sequence and create a series, then we write

which can be written, in general, as

The proof for equation [link] can be found under the Advanced block that follows:

Derivation of the finite squared series

We will now prove the formula for the finite squared series:

We start off with the expansion of ${(k+1)}^3$ .

If we add all the terms on the right and left, we arrive at

Therefore,

Finite geometric series

When we sum a known number of terms in a geometric sequence, we get a finite geometric series . We can write out each term of a geometric sequence in the general form:

where

• $n$ is the index of the sequence;
• $a_n$ is the $n^{th}$ -term of the sequence;
• $a_1$ is the first term;
• $r$ is the common ratio (the ratio of any term to the previous term).

By simply adding together the first $n$ terms, we are actually writing out the series

We may multiply the above equation by $r$ on both sides, giving us

You may notice that all the terms on the right side of [link] and [link] are the same, except the first and last terms. If we subtract [link] from [link] , we are left with just

Dividing by $(r-1)$ on both sides, we arrive at the general form of a geometric series:

The following video summarises what you have learnt so far about sequences and series:

Exercises

1. Prove that
   $$a+ar+a{r}^2+...+a{r}^{n-1}=\frac{a(1-r^n)}{(1-r)}$$
2. Find the sum of the first 11 terms of the geometric series $6+3+\frac{3}{2}+\frac{3}{4}+...$
3. Show that the sum of the first $n$ terms of the geometric series
   $$54+18+6+...+5{\left(\frac{1}{3}\right)}^{n-1}$$
   is given by $81-3^{4-n}$ .
4. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the first 7 terms.
5. Solve for $n$ : ${\sum }_{t=1}^{n}8{\left(\frac{1}{2}\right)}^t=15\frac{3}{4}$ .
6. The ratio between the sum of the first three terms of a geometric series and the sum of the 4 ^th -, 5 ^th - and 6 ^th -terms of the same series is $8:27$ . Determine the common ratio and the first 2 terms if the third term is 8.
7. Given the geometric sequence $1;-3;9;\cdots$ determine:
   1. The 8 ${}^{\mathrm{th}}$ term of the sequence
   2. The sum of the first 8 terms of the sequence.
8. Determine:
   $$\sum _{n=1}^{4}3·2^{n-1}$$

Infinite series

Thus far we have been working only with finite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the first $n$ terms. In this section, we consider what happens when we add infinitely many terms together. You might think that this is a silly question - surely the answer will be $\infty$ when one sums infinitely many numbers, no matter how small they are? The surprising answer is that while in some cases one will reach $\infty$ (like when you try to add all the positive integers together), there are some cases one will get a finite answer. If you don't believe this, try doing the following sum, a geometric series, on your calculator or computer: