How Do You Sum a Series Using Sigma Notation and Identities?

AI Thread Summary
The discussion focuses on summing the series (5^2) + (11^2) + (17^2) +...+ (18n-1)^2 using sigma notation. The initial attempt at expressing the series in sigma notation as ∑(6n-1)^2 was found to be incorrect for solving part b. Participants clarified that the correct bounds for n should be from 1 to 3k, which aligns with the requirement to match the sum to the identity 3n(108n^2 + 36n + 1). The conversation highlights the importance of correctly identifying the bounds for the sigma notation to facilitate the solution. Understanding these bounds is crucial for successfully applying sigma identities to solve the series.
Liquidxlax
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Homework Statement


(5^2) + (11^2) + (17^2) +...+ (18n-1)^2

a)Write the sum in sigma notation
b)Using the sigma identities solve the sum (easy to do)

Homework Equations



∑i = .5*k*(k+1) etc

The Attempt at a Solution



The problem I'm having is with the 25 and 121. I thought it was

∑ (6n-1)^2

where n goes from 1 to k,
but I noticed that this does not work for part b then.
I haven't done induction in 4 years, so unfortunately I forget.

Unless it isn't from 1 to k, but from 1 to 3k
 
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Liquidxlax said:

Homework Statement


(5^2) + (11^2) + (17^2) +...+ (18n-1)^2

a)Write the sum in sigma notation
b)Using the sigma identities solve the sum (easy to do)


Homework Equations



∑i = .5*k*(k+1) etc

The Attempt at a Solution



The problem I'm having is with the 25 and 121. I thought it was

∑ (6n-1)^2

where n goes from 1 to k,
but I noticed that this does not work for part b then.
I haven't done induction in 4 years, so unfortunately I forget.

Unless it isn't from 1 to k, but from 1 to 3k
How about each term is of the form (6k - 1)^2 ...

Then k has k go from 1 to ___ . (Fill in the blank.)
 
Liquidxlax said:
∑ (6n-1)^2
where n goes from 1 to k,
Yes
but I noticed that this does not work for part b then.
It doesn't? I don't see your difficulty. Please post your working up to where you're stuck.
 
haruspex said:
Yes
It doesn't? I don't see your difficulty. Please post your working up to where you're stuck.

yeah it does go from 1 to 3k, not 1 to k because it needs to be for part b equal to

3n(108n^2 +36n +1)

Where i was stuck was knowing the bounds
 

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