Mathematical induction w/ Summation question

AI Thread Summary
The discussion centers on proving the summation formula for i(i + 1) from i = 2 to n-1 using mathematical induction. The initial steps include verifying the base case P(2) and formulating P(k) and P(k+1). A key point raised is the need to establish that S(3) = P(3) as the smallest integer for which the summation is valid. The participant acknowledges confusion regarding the induction process and plans to seek additional resources for clarification. The conversation emphasizes the importance of correctly defining the summation limits and the formula to be proven.
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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