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Given the recurrence$$T_n = 2T_{n-1}-T_{n-2},$$$$T_0=0$$$$T_1=1$$Prove by induction, that $T_n = n$.

I have the first few steps worked out.

  1. Basis: $n = 1$$$T_1=1=n=1$$

  2. Assume true for $n = k+1$$$T_{k+1}=2T_k-T_{k-1}$$

  3. We know that $T_k=k$$$T_{k+1}=2(k)-T_{k-1}$$

But where do I go from here? I don't have the value for $T_{k-1}$, so how to I continue?

Jodo1992
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    $T_{k-1} = k-1$ by the induction hypothesis. – JJC94 May 06 '16 at 01:44
  • Technically there are two types of induction: simple induction (where you only use the $T_k$ statement to prove the $T_{k+1}$ statement) and strong induction where you assume the statement to be true for $1,2,3,\ldots,k$ to prove it true for $k+1$. For more see this question – Winther May 06 '16 at 01:48
  • Isn't this what I'm trying to prove? I thought I can't use things that I'm trying to prove – Jodo1992 May 06 '16 at 01:49
  • @Winther, yes I am using strong induction here. – Jodo1992 May 06 '16 at 01:49
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    You are not assuming anything about $T_{k+1}$. You know (the strong induction hypotesis) that $T_n = n$ for $n=1,2,3,\ldots,k$ so in particular it holds for $T_{k-1}$ and $T_k$ – Winther May 06 '16 at 01:50
  • It is very easy $T_{n+1}=2T_n-T_{n-1}=2n-(n-1)=n+1$ – Piquito May 06 '16 at 02:18

2 Answers2

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Note that $T_n = 2T_{n-1}-T_{n-2}$ is equivalent to $T_n -T_{n-1} = T_{n-1}-T_{n-2}$.

Thus $T_n -T_{n-1}$ is constant and equal to $T_1 -T_{0}=1$.

Therefore, $T_n=T_{n-1}+1$ and induction is very easy now.

lhf
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We have$ T_0 = 0 , T_1 = 1 . $I will prove inductively that$ T_n = n .(1)$

Let $(1) $be true at $n= k .$ Thus ,$ T_{k+1} =2T_k-T_{k-1} = 2k-(k-1) = k+1$ . So $(1)$ is also true at $n = k .$

I have completed the proof :$T_n = n$

yoursmileeeeee
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