Mathematical induction w/ Summation question

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SUMMARY

The discussion centers on proving the summation formula S(n) = sum{i*(i+1): i=2..n-1} = n(n-1)(n+1)/3 using mathematical induction. The initial steps involve verifying P(2) and establishing the induction hypothesis for P(k) and P(k+1). A critical point raised by R.G. Vickson emphasizes the need to start the induction at n=3, as it is the smallest integer for which the summation is valid. The participant acknowledges a misunderstanding and expresses intent to seek additional resources for clarification.

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Jim01
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Homework Statement



Summation of i(i + 1) (with i going from i = 2 to i = n-1) = n(n-1)(n=1) / 3

a. Write P(2). Is P(2) true?

b. Write P(k)

c. Write P(k+1)

d. Prove by mathematical induction that the formula holds true for all integers
n \geq 2



Homework Equations



N/A

The Attempt at a Solution



a. P(2): i(i+ 1) + ... + (n-1)[(n-1)+1] = n(n-1)(n+1) / 3

= (2-1)[(2-1) + 1] = 2(2-1)(2 + 1) / 3

P(2): 2 = 2

P(2) is true

b. P(k): ...+ (k-1)[(k-1)+1] = k(k-1)(k+1) / 3

P(k) = k(k-1) = k(k2 - 1)/3

c. P(k+1): (k+1)(k-1) = (k+1)[(k+1)2 - 1) / 3
= k2 - 1 = k3 + 3k2 + 2k / 3

d. Left-Hand side of P(k+1) = i(i + 1) + ... + (k+1)(k-1)
= i(i + 1) + ... + k(k-1) + (k+1)(k-1)
= k(k2 - 1)/3 + 3k2 -3 / 3
= 4k2 - 4


Right-Hand side of P(k+1) = k3 + 3k2 + 2k / 3

4k2 - 4 = k3 + 3k2 + 2k / 3


I've went over this several times and it doesn't work out so I am obviously doing something
wrong, but I am not sure where I am making the mistake.
 
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You are doing induction wrongly; perhaps you do not understand what is involved. To clarify: let S(n) = sum{i*(i+1): i=2..n-1} and let P(n) = n*(n-1)*(n+1)/3. You are being asked to prove that S(n) = P(n) for all n >= 3. To show it by induction, you must establish that S(3) = P(3) (n = 3 is the smallest integer for which the summation S(n) makes sense---unless the lower limit is i = 1 instead of i = 2). Anyway, having established that S(3) = P(3) and *assuming that* S(k) = P(k) (for some k >= 3), then you must prove that S(k+1) = P(k+1) also holds. That will prove that S(n) = P(n) holds for all n >= 3.

R.G. Vickson
 
Ray Vickson said:
You are doing induction wrongly; perhaps you do not understand what is involved. To clarify: let S(n) = sum{i*(i+1): i=2..n-1} and let P(n) = n*(n-1)*(n+1)/3. You are being asked to prove that S(n) = P(n) for all n >= 3. To show it by induction, you must establish that S(3) = P(3) (n = 3 is the smallest integer for which the summation S(n) makes sense---unless the lower limit is i = 1 instead of i = 2). Anyway, having established that S(3) = P(3) and *assuming that* S(k) = P(k) (for some k >= 3), then you must prove that S(k+1) = P(k+1) also holds. That will prove that S(n) = P(n) holds for all n >= 3.

R.G. Vickson

Thank you Mr. Vickson. You are right, I don't understand. I thought I was following the example in the book step-by-step. I will go back and reread that section. The book only gives one example, so I will take a look around the Internet and see if I can't find more examples that may shed more light on the subject.

Thank you for your help and guidance.
 
Jim01 said:
Thank you Mr. Vickson. You are right, I don't understand. I thought I was following the example in the book step-by-step. I will go back and reread that section. The book only gives one example, so I will take a look around the Internet and see if I can't find more examples that may shed more light on the subject.

Thank you for your help and guidance.
One problem is that you are being asked to prove an incorrect result (or else you made a typo when you wrote out the question). Try proving that
sum{i*(i+1), i=1..n-1} = n*(n-1)*(n+1)/3 for n >= 2.

RGV
 

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