Why is this true?
$$n=1: a_2=\frac{a_0}{2!}$$
$$n=2: a_3=0$$
$$n=3: a_4=\frac{a_0}{2^22!}$$
$$n=4: a_5=0$$
$$n=5: a_6=\frac{a_0}{2^32!}$$

Why is it
$$a_0\sum_{k=0}^{\infty} \frac{x^{2k}}{2^kk!}$$
and not
$$a_0\sum_{k=0}^{\infty} \frac{1}{2^kk!}$$

isn't it:
$$\frac{a_0}{2!}+\frac{a_0}{2^2(2!)}+\frac{a_0}{2^3(3!)}$$
$$=a_0 \left(\frac{1}{2!}+\frac{1}{2^2(2!)}+\frac{1}{2^3(3!)}$$

Question

Why is this true?
$$n=1: a_2=\frac{a_0}{2!}$$
$$n=2: a_3=0$$
$$n=3: a_4=\frac{a_0}{2^22!}$$
$$n=4: a_5=0$$
$$n=5: a_6=\frac{a_0}{2^32!}$$

Why is it
$$a_0\sum_{k=0}^{\infty} \frac{x^{2k}}{2^kk!}$$
and not 
$$a_0\sum_{k=0}^{\infty} \frac{1}{2^kk!}$$

isn't it:
$$\frac{a_0}{2!}+\frac{a_0}{2^2(2!)}+\frac{a_0}{2^3(3!)}$$
$$=a_0 \left(\frac{1}{2!}+\frac{1}{2^2(2!)}+\frac{1}{2^3(3!)}$$

anonymous · Answer

$$=a_0 \left(\frac{1}{2!}+\frac{1}{2^2(2!)}+\frac{1}{2^3(3!)}\right)$$

error on n=5....should be: 2^3 (3!) in the denominator

anonymous · Answer

@experimentX

anonymous · Answer

@Zarkon

experimentx · Answer

could you rectify that latex error

anonymous · Answer

yep it's in my second post the one with =a_0(....)

experimentx · Answer

|dw:1355853950238:dw|
from the data you have given i could say ^^