How do I write a series in sigma notation?

Question

Consider the sum: (5)² + (11)² + (17)² + ... + (18n - 1)²

 

How do I write this in sigma notation? The final term is really confusing me.

 

Answer 1

Here are two methods for solution
Method 1:
Let's start with the final term: (18n -1)2
Now let's rerwite the other terms with respect to the final term:
(5)² = (18 * n - 1)² => 5 = 18n - 1 => n = 1/3
(11)² = (18 * n - 1)² => 11 = 18 * n - 1 => n = 2/3
(17)² = (18 * n - 1)² => 17 = 18 * n - 1 = > n = 3/3

Take the terms to be a sequence of their own, a_i, where:

a₁ = 18(1/3) - 1

a₂ = 18(2/3) - 1

a₃ = 18(3/3) - 1

...

a_i = 18(i/3) - 1

So we have:
∑i(18i/3 - 1)²

Method 2:
We have (5)², (11)², (17)², ... (18n - 1)² as the terms of our sequence.
We may simplify our sequence by just dealing with the changing parts (removing the squares):
5, 11, 17, ..., 18n - 1
So we have a sequence, ai, where:
a1 = 5
a2 = 11 = a1 + 6
a3 = 17 = a2 + 6 = (a1 + 6) + 6
...
a_i>1 = ai-1 + 6 = ai-1 + 18/3
Then:
a_i = 5 + (i - 1) * (18/3)
a_i = 5 + 18i/3 - 18/3
a_i = 18i/3 - 1
So we have:
∑_i(18i/3 - 1)²

Answer 2

In this case, 'n' is going to have to be replaced with another variable (let's use k).

 

If you consider the sum, you may notice that each term increases by "6". For example, you have 5², 11² and 17² (17 is 6 greater than 11, which is 6 greater than 5). Using simple algebra, set 18n-1=5. You will find that n=1/3. Do the same for 8n-1=11 and 8n-1=17 and n=2/3 and 3/3, respectively. Now you can see that each term increases by a factor of 18(k/3)-1.

 

So your sigma notation will be:  ∑ (18(k/3)-1)² from k=1 to k=3n. (I can't format it correctly on here but the k=1 would be underneath the sigma and the k=3n would be above the sigma.

 

Hope this helped!

Answer 3

(5)² + (11)² + (17)² + ... + (18n - 1)² = ∑_k=1³ⁿ (6k-1)²