Mathematical induction exercise

[danielr96]

I didnt know where to post this so, Im sorry if I have chosen the wrong forum.

\(\displaystyle a^n\, -\, b^n\, =\, (a\, -\, b)\, \displaystyle{\sum_{k=1}^n\, a^{n-k}\, b^{k-1}}\)

I cant prove that p(n) implies p(n+1)

 

[Ishuda]

I didnt know where to post this so, Im sorry if I have chosen the wrong forum.

\(\displaystyle a^n\, -\, b^n\, =\, (a\, -\, b)\, \displaystyle{\sum_{k=1}^n\, a^{n-k}\, b^{k-1}}\)

I cant prove that p(n) implies p(n+1)

Assuming p(n), try taking the sum for p(n+1) up to the nth term [Hint, factor out an a] and adding the last term by itself. What does that give you?

 

[danielr96]

I tried this

a^n + b^n + (a-b).a^(n+1-(n+1)).b^n+1-k = (a-b).sum(a^(n-k).b^(k-1))+(a-b).a^(n+1-(n+1)).b^n+1-k

 

[Ishuda]

I tried this

a^n + b^n + (a-b).a^(n+1-(n+1)).b^n+1-k = (a-b).sum(a^(n-k).b^(k-1))+(a-b).a^(n+1-(n+1)).b^n+1-k

Try
\(\displaystyle (a - b) \sum_{k=1}^{k=n+1}a^{n+1-k} b^{k-1} \) = \(\displaystyle (a - b) (\sum_{k=1}^{k=n}a^{n+1-k} b^{k-1}\) + a⁰ bⁿ)

Now a⁰ is 1 and you can factor out an a from the right hand summation. Now distribute the (a-b) multiply through and go from there [Hint: look at what p(n) says]

 

[danielr96]

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I cant find whats wrong there...

 

[Ishuda]

​<link removed>

I cant find whats wrong there...

Why do you do the second line; aⁿ - bⁿ + (a - b) a⁰ bⁿ is not equal to aⁿ⁺¹ - bⁿ⁺¹

Try writing aⁿ⁺¹ - bⁿ⁺¹ = a (aⁿ - bⁿ) + (a-b) bⁿ

or, if you really get stuck, highlight the following between the two lines of *'s
*****************************************
Lets start with
(a-b)\(\displaystyle \sum_{k=1}^{k=n+1} a^{n+1-k} b^{k-1}\) = (a-b) ( \(\displaystyle \sum_{k=1}^{k=n} a^{n+1-k} b^{k-1}\) + bⁿ )
= (a-b) \(\displaystyle \sum_{k=1}^{k=n} a^{n+1-k} b^{k-1}\) + (a-b) bⁿ
= a (a-b) ( \(\displaystyle \sum_{k=1}^{k=n} a^{n-k} b^{k-1}\) + (a-b) bⁿ
= a (aⁿ - bⁿ) + (a-b) bⁿ
= aⁿ⁺¹ - a bⁿ + a bⁿ - bⁿ⁺¹
= aⁿ⁺¹ - bⁿ⁺¹
*****************************************