Question

So I have this function: \[f(x)=(1+x)^{-3}\] And I need to find the Maclaurin's serie. I have found the following: \[=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)x^2}{2!}+...\] \[=1-\frac{3x}{1!}+\frac{12x^2}{2!}-\frac{60x^3}{3!}+...\] How do I convert this to a Sum?

Answer 1

I will rely on your result,

\(\large\color{#000000 }{ \displaystyle \sum_{k=0}^{\infty} \frac{}{k!}}\)

but up to here it is fairly obvious if you look at the denominators,
and recall also that 0!=1.

Answer 2

Something happened to the equation in your question,
but I can still read it...

Answer 3

wait, is there a reason you removed it?
you think it's incorrect?

Answer 4

I havent removed anything?

Answer 5

oh you didn't remove it and my latex is just not readin

Answer 6

I think it is bugging. But I wrote:
So I have this function: \[f(x)=(1+x)^{-3}\] And I need to find the Maclaurin's serie. I have found the following: \[=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)x^2}{2!}+...\] \[=1-\frac{3x}{1!}+\frac{12x^2}{2!}-\frac{60x^3}{3!}+...\] How do I convert this to a Sum?

Answer 7

Yea, so I am having trouble finding the nominator.

Answer 8

I missed it, I will erase the mess I made.

Answer 9

\[\sum_{k=0}^{\infty}(-1)^k \frac{ (k+1)(k+2) }{ 2 }\]

Answer 10

Wouldnt that make sense?

Answer 11

missed \[x^n\] at the end

Answer 12

\[\sum_{k=0}^{\infty}(-1)^k \frac{ (k+1)(k+2) }{ 2 }x^k\]

Answer 13

and without k! ?

Answer 14

Yea, but maybe this isnt a maclaurins serie?

Answer 15

I will see if I can redo it and rethink it...

Answer 16

I am lagging I need more time because that; apologize.

Answer 17

Dont worry

Answer 18

I am lagging so bad I am sorry

Answer 19

No worries, ill take a look at some other questions. Thanks for your help though :)

Answer 20

\[\sum_{k=0}^{\infty}(-1)^k \frac{ (k+1)(k+2) }{ 2 }x^k\]

was correct

Answer 21

Sorry for wasting your time....
(a bit frusturating with this internet.
I got to go offline)

Answer 22

No worries, have a great day