Infinite series -- is this an acceptable answer?

[KrabLord]

The problem is: Find the function which equals
[MATH] y=1-2x+4x^2- ... [/MATH]
Thus, my reasoning was as follows:
[MATH] F = \frac{1}{(1+x)^2} = 1-2x+3x^2-... \implies xF = x-2x^2+3x^3-... \implies 1-2xF = 1-2x+4x^2-...[/MATH]Furthermore, [MATH] 1-2xF = 1 - \frac{2x}{(1+x)^2} = \frac{1+x^2}{(1+x)^2} = y[/MATH]
However, the book (Calculus by Gilbert Strang, first edition) offers a far more parsimonious solution:
[MATH]( \frac{1}{1-x} = 1+x+x^2+x^3+... ) \land ( x = -2x) \implies \frac{1}{1+2x} = 1-2x+4x^2-... [/MATH]
So, my question is, is my answer also correct? If not, why?
Also, here are their graphs on Desmos:

 

[Deleted member 4993]

The problem is: Find the function which equals
[MATH] y=1-2x+4x^2- ... [/MATH]
Thus, my reasoning was as follows:
[MATH] F = \frac{1}{(1+x)^2} = 1-2x+3x^2-... \implies xF = x-2x^2+3x^3-... \implies 1-2xF = 1-2x+4x^2-...[/MATH]Furthermore, [MATH] 1-2xF = 1 - \frac{2x}{(1+x)^2} = \frac{1+x^2}{(1+x)^2} = y[/MATH]
However, the book (Calculus by Gilbert Strang, first edition) offers a far more parsimonious solution:
[MATH]( \frac{1}{1-x} = 1+x+x^2+x^3+... ) \land ( x = -2x) \implies \frac{1}{1+2x} = 1-2x+4x^2-... [/MATH]
So, my question is, is my answer also correct? If not, why?
Also, here are their graphs on Desmos: View attachment 16925

You write:

[MATH] F = \frac{1}{(1+x)^2} = 1-2x+3x^2-...[/MATH]
How did you derive that?

 

[KrabLord]

You write:

[MATH] F = \frac{1}{(1+x)^2} = 1-2x+3x^2-...[MATH] How did you drive that?[/MATH][/MATH]

[MATH] Q=\frac{-1}{1+x} = -1+x-x^2+x^3-... \implies \frac{dQ}{dx} = \frac{1}{(1+x)^2} = 1-2x+3x^2-... [/MATH]
Is that sufficient?

 

[Dr.Peterson]

The problem is: Find the function which equals
[MATH] y=1-2x+4x^2- ... [/MATH]
Thus, my reasoning was as follows:
[MATH] F = \frac{1}{(1+x)^2} = 1-2x+3x^2-... \implies xF = x-2x^2+3x^3-... \implies 1-2xF = 1-2x+4x^2-...[/MATH]Furthermore, [MATH] 1-2xF = 1 - \frac{2x}{(1+x)^2} = \frac{1+x^2}{(1+x)^2} = y[/MATH]
However, the book (Calculus by Gilbert Strang, first edition) offers a far more parsimonious solution:
[MATH]( \frac{1}{1-x} = 1+x+x^2+x^3+... ) \land ( x = -2x) \implies \frac{1}{1+2x} = 1-2x+4x^2-... [/MATH]
So, my question is, is my answer also correct? If not, why?

The question is ambiguous. We strongly recommend against defining a sequence by giving only the first few terms, and the author has done just that!

You are assuming the series is [MATH] y=1-2x+4x^2- 6x^3 + 8x^4 - ... [/MATH]; at least that is what your series is when you extend it.

It is evidently intended to be [MATH] y=1-2x+4x^2- 8x^3 + 16x^4 - ... [/MATH]. And this pattern does make more sense.

You fell into the trap of not defining the general term of either the given series or the one you started with, but working with just a few terms.

 

[KrabLord]

The question is ambiguous. We strongly recommend against defining a sequence by giving only the first few terms, and the author has done just that!

You are assuming the series is [MATH] y=1-2x+4x^2- 6x^3 + 8x^4 - ... [/MATH]; at least that is what your series is when you extend it.

It is evidently intended to be [MATH] y=1-2x+4x^2- 8x^3 + 16x^4 - ... [/MATH]. And this pattern does make more sense.

You fell into the trap of not defining the general term of either the given series or the one you started with, but working with just a few terms.

Thank you! I was frustrated that Dr. Strang only gave three terms... but I like that you extended the series to show they are not equal.