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Question:

For positive integers $n$, let the numbers $c(n)$ be determined by the rules $c(1)=1$, $c(2n)=c(n)$, and $c(2n+1)=(-1)^nc(n)$. Find the value of $\displaystyle\sum_{n=1}^{2013}c(n)c(n+2)$.

To solve this question, I tried to figure out some patterns.

we are given $c(1)=1$ but observe that if $n=1$ then $C(2)=C(1)$, which means $C(2)=1$ as well.

Also, $C(4)=C(2)=1$ and $C(8)=C(4)=1$ and so on such as $C(16)=C(8)$...

Then I tried working with the third rule which is $c(2n+1)=(-1)^n c(n)$ and i notice if $n=1$ then $c(3)=(-1)^{1} c(1)=-1$. However if $n=2$ then $c(2(2)+1)=c(5)=(-1)^2 c(2)=1$ So this shows that not every odd number will be $-1$

So I thought it could be alternate, and I tested the third rule with $n=3$ which would result in $c(7)=(-1)^3 c(3)=1$ however i get another positive $1$.

then i tried one more time with $n=4$ which is $c(9)=(-1)^4 c(4)=1$ but i get $1$ again.

So anyone have any ideas on how i can solve this problem, the pattern doesn't seem so obvious to me with $c(3)$ being negative.

Mario Carneiro
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Caddy Heron
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1 Answers1

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Rewrite the sum as

$$S = \sum_{m=1}^{1006}c(2m)c(2m+2) + \sum_{m=0}^{1006}c(2m+1)c(2m+3).$$

The first sum covers the even $n$; the second covers the odd $n$.

Then replace:

$$S = \sum_{m=1}^{1006}c(m)c(m+1) + \sum_{m=0}^{1006}(-1)^m c(m) (-1)^{m+1} c(m+1)$$

$$S = c(1)c(3) + \sum_{m=1}^{1006} [1 + (-1)^m(-1)^{m+1}] c(m)c(m+1).$$

Since $(-1)^m(-1)^{m+1} = (-1)^{2m+1} = -1$ for all integer $m$, every term in the sum vanishes and we're left with

$$S = c(1)c(3) = -1.$$

John
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