# Manipulating series to find the recursive formula

Ok so I am stuck. I need to get all the $n$'s to $=0$ but I can't reduce my series which has $n=2$ to $0$ because then I will have undone all my work in the first place to get all the $X^n$'s to the same power.. I am really confused on what to do.. Here is where I'm at now:

$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + \sum ^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - 6\sum ^{\infty}_{n=0} na_nX^n + 10 \sum ^{\infty}_{n=0}a_nX^n$$

any guidance would be great. Thx in advance

• It is much easier to get all the n's to 2 – user174622 Apr 26 '15 at 2:08
• That wouldn't make any sense, because then I wouldnt have my X's to the same power. – Yusha Apr 26 '15 at 2:09
• Why would you have to change the powers? – user174622 Apr 26 '15 at 2:12
• What do you mean? You can't just change $n =0$ to $n=2$ without adding $n-2$ to everywhere there is a $n$ right?? – Yusha Apr 26 '15 at 2:14
• The powers are all n so it doesn't matter. – user174622 Apr 26 '15 at 2:15

$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + (2a_2+6a_3+\sum ^{\infty}_{n=2} (n+2)(n+1)a_{n+2}X^n) - (a_1+6\sum ^{\infty}_{n=2} na_nX^n) + (a_1+10\sum ^{\infty}_{n=2}a_nX^n)$$

$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_nX^n + (n+2)(n+1)a_{n+2}X^n) - (6 na_nX^n) + (10a_nX^n)]$$

$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_n + (n+2)(n+1)a_{n+2}) - (6 na_n) + (10a_n)]X^n$$

$$(2a_2+6a_3)=0\rightarrow a_2=-3a_3$$
$$\sum ^{\infty}_{n=2} [(n^2-7n+10))a_n + (n+2)(n+1)a_{n+2}) ]X^n=0\sum ^{\infty}_{n=2} [(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2}) ]X^n=0$$
$$(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2})=0\rightarrow a_{n+2}=-\frac{(n-5)(n-2)}{(n+2)(n+1)}a_n$$
• shouldn;t you have $a_0+a_1 + 10$? – Yusha Apr 26 '15 at 2:32