I'm a bit of a maths noob so please bear with me with what is probably a really dumb question, but I could really do with some help - I'm self-learning at home.

I'm stuck on the question below from Discrete Mathematics for Computing (Peter Grossman) on proving a recursive definition by induction. I understand everything up to the highlighted line in yellow.

It asks to create a non-recursive version of the formula

$t(n) = t(n-1) + 2n - 1$

Writing down the first few terms indicates a squares sequence, so a non-recursive form is:

$t(n) = n^2$

The question asks to then prove this is correct. The base is shown in the question, t(1) = 1 so the base is:

$t(1) = 1^2$

I then assume (inductive hypothesis) that

$t(k) = k^2$

I then need to prove that it's true for every successive term, i.e. with n = k+1. So I thought the next step would be:

$t(k+1) = (k+1)^2$

But instead the book jumps back to the original, non-recursive formula at this point (yellow line), and I just don't get that bit - surely we are now proving t(n) - t(n-1) + 2n -1 instead of t(n) = n^2? I can't assume these are equivalent at this point, as we are only conjecturing that they are the same... confused!


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To prove the inductive step we have to

Use the assumption that $t(k) = k^2$ to prove that $t(k + 1) = (k+1)^2$.

The aim is to prove that $t(k + 1) = (k+1)^2$, but we can't just write this down; we have to derive it from the assumption that $t(k) = k^2$.

By definition, $t(n) = t(n-1) + (2n-1)$, and hence $$t(k+1) = t(k) + 2(k+1)-1$$ We then use the assumption that $t(k) = k^2$ to show that $$t(k+1) = k^2 + 2(k+1) - 1 = k^2 + 2k + 1$$

From here we can see that $$t(k+1) = (k+1)^2$$

Hence, using our assumption that the result is true for $k$, we have shown it is true for $k+1$. Since it is true for $k=1$, it is therefore true for all $k$ by induction.

  • $\begingroup$ Okay thanks, I think I get it now - think my confusion was not realising it's okay to assume that the first equation t(n) = t(n-1)+2n-1 IS correct and always true, and so you can use it in proving the second equation t(n) = n^2. The rest makes sense. Think I just need practice. Thank you $\endgroup$ May 12 '15 at 15:03
  • $\begingroup$ The idea is to show that given a definition of $t(n)$, there is an alternative definition given by $t(n)=n^2$, so you can certainly assume the first equation. $\endgroup$
    – Mathmo123
    May 12 '15 at 17:10

The point is to prove that $t(k)=k^2$ implies $t(k+1)=(k+1)^2$.


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