`f(x)=1/(1+x)^4` Use the binomial series to find the Maclaurin series for the function.

Recall binomial series  that is convergent when `|x|lt1` follows: 

`(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(n!)`

or`(1+x)^k= 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4-` ...

 For the given function `f(x) =1/(1+x)^4` , we may  apply Law of Exponents: `1/x^n = x^(-n)` to rewrite it as:

`f(x) = (1+x)^(-4)`

This now resembles `(1+x)^k` for binomial series.  

By comparing "`(1+x)^k` " with "`(1+x)^(-4)` ", we have the corresponding values:

`x=x` and `k = -4` .

 Plug-in the values  on...

Recall binomial series  that is convergent when `|x|lt1` follows: 

`(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(n!)`

or`(1+x)^k= 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4-` ...

 For the given function `f(x) =1/(1+x)^4` , we may  apply Law of Exponents: `1/x^n = x^(-n)` to rewrite it as:

`f(x) = (1+x)^(-4)`

This now resembles `(1+x)^k` for binomial series.  

By comparing "`(1+x)^k` " with "`(1+x)^(-4)` ", we have the corresponding values:

`x=x` and `k = -4` .

 Plug-in the values  on the formula for binomial series, we get:

`(1+x)^(-4)=sum_(n=0)^oo ((-4)(-4-1)(-4-2)...(-4-n+1))/(n!)x^n`

               `= 1 + (-4)x + ((-4)(-4-1))/(2!) x^2 + ((-4)(-4-1)(-4-2))/(3!)x^3 +((-4)(-4-1)(-4-2)(-4-3))/(4!) x^4-` ...

` = 1 + (-4)x + ((-4)(-5))/(2!) x^2 + ((-4)(-5)(-6))/(3!)x^3 +((-4)(-5)(-6)(-7))/(4!) x^4-` ...

` = 1 -4x + 20/(2!) x^2 -120/(3!)x^3 +840/(4!)x^4-` ...

` = 1- 4x +10x^2 -20x^3 +35x^4-` ...

Therefore, the Maclaurin series  for  the function `f(x) =1/(1+x)^4` can be expressed as:

`1/(1+x)^4 =1- 4x +10x^2 -20x^3 +35x^4-` ...