`f(x)=1/(1+x)^2` Use the binomial series to find the Maclaurin series for the function.

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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!)x^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 given function `f(x) =1/(1+x)^2` , we may  apply Law of Exponents: `1/x^n = x^(-n)` to rewrite it as:

`f(x) = (1+x)^(-2)`

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

By comparing "`(1+x)^k` " with "`(1+x)^(-2)` ", we have the corresponding values:

`x=x ` and `k = -2` .

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

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

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

             `= 1 -2x + 6/(2!) x^2 -24/(3!)x^3 +120/(4!)x^4-` ...

              `= 1- 2x +3x^2 -4x^3 +5x^4-` ...

              or  `sum_(n=0)^oo (-1)^n (n+1)x^n`

Therefore, the Maclaurin series  for  the function `f(x) =1/(1+x)^2` can be expressed as:

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


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

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