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

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SUMMARY

The discussion focuses on summing a series expressed in sigma notation, specifically the series (5^2) + (11^2) + (17^2) +...+ (18n-1)^2. The correct sigma notation is identified as ∑ (6n-1)^2, where n ranges from 1 to 3k. The participants clarify that the bounds for n must be adjusted to ensure the sum equals 3n(108n^2 + 36n + 1) for part b of the problem. The use of sigma identities is emphasized as a straightforward method to solve the sum.

PREREQUISITES
  • Understanding of sigma notation and summation
  • Familiarity with polynomial identities
  • Knowledge of mathematical induction
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study sigma notation and its applications in series
  • Learn about polynomial identities and their proofs
  • Review mathematical induction techniques
  • Practice solving summation problems using sigma identities
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Students in mathematics, particularly those studying series and sequences, educators teaching algebraic concepts, and anyone looking to enhance their understanding of sigma notation and summation techniques.

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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