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So I have a recursive definition for a sequence, which goes as follows:

$$s_0 = 1$$ $$s_1 = 2$$ $$s_n = 2s_{n-1} - s_{n-2} + 1$$

and I have to prove the following proposition: The $n$th term of the sequence defined above is $s_n = \frac{n(n+1)}{2} + 1$.

To prove this, I started it off by induction. The base case is for $n = 0$ and its true since the non-recursive and recursive results match. I assumed the following hypothesis to be true for some $n = k$: that the $n$th term of the sequence is $s_k = \frac{k(k+1)}{2} + 1$. Then, in the induction step, we need to show that $s_{k + 1} = \frac{(k+1)(k+2)}{2} + 1$.

Using the recursive definition of the sequence, we have that $s_{k+1} = 2s_k - s_{k-1} + 1$, so we can use the hypothesis to replace $s_k$ and $s_{k-1}$ by their non-recursive formulas:

$$ s_{k+1} = 2 (\frac{k(k + 1)}{2} + 1) - (\frac{(k-1)k}{2} + 1) + 1$$

After simplifying i get

$$ s_{k+1} = \frac{k(k+3)}{2} + 2 $$

which is clearly wrong. Can someone point out what I'm doing wrong and where I can go with this?

EDIT: The answer I have given is correct, except that we need to simplify further to get the form we want:

$$ \frac{k(k+3)}{2} + 2 = \frac{k^2 + 3k + 4}{2}$$

after expansion and common denominators, and then this is clearly equal to

$$ \frac{(k+1)(k+2)}{2} + 1 = \frac{k^2 + 3k + 2}{2} + 1 = \frac{k^2 + 3k + 4}{2} $$

noobProgrammer
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  • Your induction hypothesis should be that it works for all $k<n$, which isn't what you did, but acted as if. – Git Gud Sep 14 '13 at 23:00
  • As I mention in my answer, you left out the $+1$ when you applied the recursion to get $s_{k+1}$. Of course, somehow the $+1$ snuck back in to $\frac{k(k+3)}{2}+2$. Furthermore, you want to get $s_{k+1}=\frac{(k+1)(k+2)}{2}+1$, which is what you got. – robjohn Sep 14 '13 at 23:37

2 Answers2

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Your work is fine.

$${s_{k + 1}} = \frac{{k(k + 3)}}{2} + 2 = \frac{{k(k + 3) + 2}}{2} + 1 = \frac{{{k^2} + 3k + 2}}{2} + 1 = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2} + 1$$

As GitGud has pointed out, you should assume that the formulas for $s_k$ and $s_{k-1}$ hold true, since you need them both for your recurrence.

Pedro
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  • I had sometimes students in the past erasing good work because they thought the final answer was wrong. They didn't realize their answer was the right one, written in an equivalent form.... – N. S. Sep 14 '13 at 23:01
  • @N.S. Tragic. Lack of self confidence? Lack of knowledge? – Pedro Sep 14 '13 at 23:02
  • I think lack of self confidence and lack of experience... – N. S. Sep 14 '13 at 23:06
  • Thanks Peter. I did assume them true for my work but apparently I just suck at equivalent forms of polynomials. Thanks. – noobProgrammer Sep 15 '13 at 13:27
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At the end of this answer is a brief description of inverting finite difference operators. In the case here $$ \begin{align} \Delta^2s_k &=s_n-2s_{n-1}+s_{n-2}\\ &=1 \end{align} $$ Is a second order finite difference operator. Inverting the operator by summing twice, we get that the solution is of the form $$ s_n=\frac12n^2+c_1n+c_0 $$ We can compute the constants by plugging in known values: $$ \frac12\cdot0^2+c_i\cdot0+c_0=s_0=1\\ \frac12\cdot1^2+c_i\cdot1+c_0=s_1=2 $$ Thus, $c_0=1$ and $c_1=\frac12$. Therefore, $$ \begin{align} s_n &=\frac12n^2+\frac12n+1\\ &=\frac{n(n+1)}{2}+1 \end{align} $$

The only problem in your solution is that you left out the $+1$ $$ \begin{align} s_{k+1} &=2\left(\frac{k(k+1)}{2}+1\right)-\left(\frac{(k-1)k}{2}+1\right)\color{#C00000}{+1}\\ &=(k^2+k+2)-\frac12(k^2-k+2)+1\\ &=\frac{k^2+3k+4}{2}\\ &=\frac{(k+1)(k+2)}{2}+1 \end{align} $$ However, since the answer you got was $\frac{k(k+3)}{2}+2$, perhaps the omission of the $+1$ was simply a typo. In that case, you got the right answer since $$ \frac{k(k+3)}{2}+2=\frac{(k+1)(k+2)}{2}+1 $$

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robjohn
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