Question

Write the following series in summation notation. 1, 4, 9, 16, 25, 36

Answer 1

Answer:

1 , 2+2, 3+3+3, 4+4+4+4 , 5+5+5+5+5 , and so on

Step-by-step explanation:

1, 4, 9, 16, 25, 36

can be written as

1x1 , 2x2, 3x3, 4x4 and so on

Answer 2

Answer:

$\sum_{k=0}^{n}2k+1$

Step-by-step explanation:

Let's take a look at how much is added from one term to the next and see if we can recognize a pattern!

• 1 to 4: +3
• 4 to 9: +5
• 9 to 16: +7
• 16 to 25: +9
• 25 to 36: +11

If we start by listing the first number in sequence, 1, we get the familiar list:

1, 3, 5, 7, 9, 11

Of course, this is simply the list of the first six odd numbers. To get the first term, we add the first 1 odd number, to get the second, we add first 2 (1 +3), to get the third, we add the first 3 (1 + 3 + 5), and so on. How do we shorten these sums so we're not writing out dozens of terms as the sequence goes on?

Sigma notation: Summing things up

For situations where we're adding up a lot of numbers that follow a predictable pattern, mathematicians came up with something called summation notation, or sigma notation, coming from the Greek S, Σ, short for "sum". Here's a simple example of a sum expressed in sigma notation:

$\sum_{n=1}^{4}n=1+2+3+4$

Let's break this down. n acts as our counter. $\sum_{n=1}$ sets it to start counting at 1, and tells us to stop counting at n = 4. The $\sum n$ bit tells us what pattern we'll be following, in this case, each step of the way we'll be adding the value of the counter.

In summation

So how do we express the sum 1 + 3 + 5 + ... in sigma notation? First, we need an expression that describes the pattern algebraically. Ever odd number is 1 away from an even number, so we can either describe our list with the expression $2k+1$ (if we start at k = 0), or $2k-1$ (if we start at k = 1). I'll choose $2k+1$ for this problem.

We want to start counting at k = 0 , and we can choose whatever we want for our stopping point. Let's call that stopping point n, for the nth term in the series. In sigma notation, we can now write this series in the form

$\sum_{k=0}^{n}2k+1$

Let's test it out for the first few values to see if it works!

• n = 0: [2(0) + 1] = 1
• n = 1:  [2{0) + 1] + [2(1)+1] = 1 + 3 = 4
• n = 2: [2{0) + 1] + [2(1) + 1] + [2(2) + 1] = 1 + 3 + 5 = 9

Checks out!