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I understand the solution of $a_{m+n}+a_{m-n}=\frac{1}{2}(a_{2m}+a_{2n}), a_1=1$, find $ a_{1995}$ and I was able to derive it myself, however, in my first attempts I conjectured something very different, and I don't see the problem in my approach:

Using $m=0, n=0$ we obtain $a_0=0$ Using $n=0$ we obtain $a_{2m} = 4a_{m}$

Up to that point, everything is fine.

Using the fact that $a_1=1$, I found $a_{2k}=2^{2k}$ (conjecture that seems to hold).

By replacing in the original question we have : $2(a_{m+n}+a_{m-n})= a_{2m}+a_{2n} = 2^{2m}+2^{2n}$.

Now if we put $n=0$, our equation becomes $2(a_{m}+a_{m})= 2^{2m} + 1$.

I replaced with $m=1995$ and obtained $a_{1995} = \frac{1}{4}(2^{2.1995} + 1)$

Which is evidently not a solution, but I don't get why I can't solve the problem in this way.

It's seems that my conjecture is false, but how could I check it and avoid running in this issue next time?

Itimethy
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1 Answers1

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Your conjecture that $a_{2k}=2^{2k}$ is false, in general. What actually holds though is $a_{2^k}=2^{2k}$, for every positive integer $k$.

Here is a proof: You have proven that $a_{2m}=4a_m$. Using this one obtains (for example, by induction) that $a_{2^k}=(2)^2a_{2^{k-1}}=2^4(a_{2^{k-2}})=2^6(a_{2^{k-3}})=\cdots=2^{2k-2}a_2=2^{2k}$.

Why is $a_{2k}=2^{2k}$ is false, in general? Well, take $m=4,n=2$. Then from the iteration relation one has $a_6+a_2=\frac{1}{2}(a_8+a_4)=40$, i.e., $a_6=40-a_2=36\neq 2^6$.

ShyamalSayak
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