Power series, how to shift the index of a summation with a powers 2n+1 I am taking an ordinary differential equation class, and we are currently learning about power series. One thing that comes up is index shifting, for the most part, I can shift the index quite easily, but in the following case, I end up with a fraction index. I have an ODE of the form y'' + 2xy' + y = 0, with one of the solutions being (in power series form):
y = $\sum_{n=0}^{\infty} b_n x^{2n+1}$
finding y' and y'', gives me:
y' = $\sum_{n=1}^{\infty} (2n+1)b_n x^{2n}$
y'' = $\sum_{n=2}^{\infty} (2n)(2n+1)b_n x^{2n-1}$
plugging it in the ODE, I get:
$\sum_{n=2}^{\infty} (2n)(2n+1)b_n x^{2n-1} + 2x\sum_{n=1}^{\infty} (2n+1)b_n x^{2n} + \sum_{n=0}^{\infty} b_n x^{2n+1} = 0$
which is simplified as
$\sum_{n=2}^{\infty} (2n)(2n+1)b_n x^{2n-1} + \sum_{n=1}^{\infty} 2(2n+1)b_n x^{2n+1} + \sum_{n=0}^{\infty} b_n x^{2n+1} = 0$
From there, I  can put y and y' together as such:
$b_0x + \sum_{n=1}^{\infty} [b_n + 2(2n+1)b_n]x^{2n+1} + \sum_{n=2}^{\infty} (2n)(2n+1)b_n x^{2n-1} = 0$
And now I'm stuck on how to shift the index so I can add both of the remaining summations together. The problem comes from having a power of 2n instead of just n, i end  with indexes that are fractions, which cannot be right. 
I tried to look around on google but every examples are showing how to shift indexes with a power series of $x^{n+a}$ and not $x^{2n+a}$
Can anyone be kind enough to show me how to shift the index for this problem (and perhaps give me another example which is similar to this)? Thank you

The right-hand sum's lowest term is $x^3$. We want $x^3$ to be of the form $x^{2n+1}$ so that we can combine the right-hand sum with the left-hand one. Clearly then, we can start the sum at n=1 and replace n with n+1 everywhere it appears in the right-hand sum. This sends 2n to 2(n+1), 2n+1 to 2n+3, and $b_n$ to $b_{n+1}$.
