Power series recurrence relations

Homework Statement

In the following series':

According to my book, we only have a common range of summation here for n >= 2.

Therefore we need to treat n = 0 and n = 1 separately.

We thus write:

[2r(r-1) + 3r - 1]c0 = 0

and

[2(r+1)r + 3(r+1)-1]c1 = 0

My question is in both cases we ignored the last (4th) series to develop these two equations. How can we just ignore this power series?

I imagine it must be because its summation begins at n=2 unlike the others, but why are we allowed to ignore it up until n = 2?

Thanks!

The Attempt at a Solution

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tiny-tim
Homework Helper
Hi IniquiTrance!

(try using the X2 and X2 tags just above the Reply box )
According to my book, we only have a common range of summation here for n >= 2.

Therefore we need to treat n = 0 and n = 1 separately.

We thus write:

[2r(r-1) + 3r - 1]c0 = 0

and

[2(r+1)r + 3(r+1)-1]c1 = 0

My question is in both cases we ignored the last (4th) series to develop these two equations. How can we just ignore this power series?

I imagine it must be because its summation begins at n=2 unlike the others, but why are we allowed to ignore it up until n = 2?

ah, but you're really writing:

([2r(r-1) + 3r - 1]c0 + 0)xr = 0

and

([2(r+1)r + 3(r+1) -1]c1 + 0)xr+1 = 0,

so you're not really ignoring the 4th sum at all, you're just not bothering to put in the zeros.

Alternatively, think of the 4th sum as starting at n = 0 like the others, and with c-1 and c-2 defined as 0

Ah, ok much clearer now. Thank you!