I'm still having trouble writing these in a short "sum" form. I have the solution, but for whatever reason I'm not able to reason my way to the solution.
$$n=0, a_2=\frac{a_0}{2!}$$
$$n=1, a_3=\frac{a_1}{3!}$$
$$n=2, a_4=\frac{a_0}{4!}$$
$$n=3, a_5=\frac{a_1}{5!}$$
$$n=4, a_6=\frac{a_0}{6!}$$
$$n=5, a_7=\frac{a_1}{7!}$$
$$n=6, a_8=\frac{a_0}{8!}$$

I understand that I will have
$$y=a_0\sum +a_1\sum$$

Question

I'm still having trouble writing these in a short "sum" form. I have the solution, but for whatever reason I'm not able to reason my way to the solution. 
$$n=0, a_2=\frac{a_0}{2!}$$
$$n=1, a_3=\frac{a_1}{3!}$$
$$n=2, a_4=\frac{a_0}{4!}$$
$$n=3, a_5=\frac{a_1}{5!}$$
$$n=4, a_6=\frac{a_0}{6!}$$
$$n=5, a_7=\frac{a_1}{7!}$$
$$n=6, a_8=\frac{a_0}{8!}$$

I understand that I will have 
$$y=a_0\sum +a_1\sum$$

anonymous · Answer

@phi

anonymous · Answer

@blues

anonymous · Answer

@Zarkon

anonymous · Answer

Originally I had
$$a_{n+2}=\frac{a_n}{(n+1)(n+2)}$$

zarkon · Answer

are you trying to write $$y=\sum_{n=0}^{\infty}a_n$$

but then to break it up into two sums?

anonymous · Answer

If one sum is sufficient we can write it in one sum, but how do I know that I need to sums or one?

zarkon · Answer

two sums would be nice since  they would be in terms of  $a_0$ and $a_1$

zarkon · Answer

if you look t the even terms...
$$a_0+a_2+a_4+a_6+\cdots$$

$$a_0+\frac{a_0}{2!}+\frac{a_0}{4!}+\frac{a_0}{6!}+\cdots$$
$$a_0\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots\right)$$
$$a_0\left(\frac{1}{0!}+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots\right)$$

$$a_0\left(\frac{1}{0!}+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots\right)=a_0\sum_{k=0}^{\infty}\frac{1}{(2k)!}$$

anonymous · Answer

makes sense! Let's see if I can do a_1

anonymous · Answer

$$a_3+a_5+a_7+a_9$$
$$\frac{a_1}{3!}+\frac{a_1}{5!}+\frac{a_1}{7!}+\frac{a_1}{9!}+\dots$$
$$a_1\left(\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{9!}+\dots\right)=a_1\sum_{k=0}^{\infty}\frac{1}{2k+1}$$

anonymous · Answer

ooops!

anonymous · Answer

$$a_1\left(\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{9!}+\dots\right)=a_1\sum_{k=0}^{\infty}\frac{1}{(2k+1)!}$$

zarkon · Answer

right...we should probably start both sums at k=1

anonymous · Answer

sure

zarkon · Answer

since looking above your n=0 corresponds to $a_2$

zarkon · Answer

$$y=a_0\sum_{k=1}^{\infty}\frac{1}{2k!}+a_1\sum_{k=1}^{\infty}\frac{1}{(2k+1)!}$$

zarkon · Answer

$$y=a_0\sum_{k=1}^{\infty}\frac{1}{(2k)!}+a_1\sum_{k=1}^{\infty}\frac{1}{(2k+1)!}$$

anonymous · Answer

Finally It makes sense. I think writing out the final step with the big parenthesis allowed me to see the sum. THanks a million!

zarkon · Answer

np