Khadija Wells

2020-10-19

a) Find the Maclaurin series for the function

$f(x)=\frac{1}{1}+x$

b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function

$g(x)=\frac{1}{(x+1{)}^{2}}$

c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function

$h(x)=\frac{1}{(x+1{)}^{3}}$

d) Find the sum of the series

$\sum _{n=3}^{\mathrm{\infty}}\frac{n(n-1)}{2n}$

This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is$\frac{7}{2}$

b) Use differentiation of power series and the result of part a) to find the Maclaurin series for the function

c) Use differentiation of power series and the result of part b) to find the Maclaurin series for the function

d) Find the sum of the series

This is a Taylor series problem, I understand parts a - c but I do not understand how to do part d where the answer is

Nathalie Redfern

Skilled2020-10-20Added 99 answers

It took me a while to get this, but finally I got it. Had to try a lot of tricks to get this series.

We have the series for$\frac{1}{(x+1{)}^{3}}$ as shown below:

$f(x)=\frac{1}{(1+x{)}^{3}}=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{2}(-1{)}^{n}(n+1)(n+2){x}^{n}$

We can similarly write a series for$\frac{1}{(1-x{)}^{3}}$ as shown below:

$f(x)=\frac{1}{(1-x{)}^{3}}=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{2}(n+1)(n+2){x}^{n}$

Now, since the index of the summation in our question starts from n=3, we could shift the index of this series to n=3. I tried that, but it didnt

We have the series for

We can similarly write a series for

Now, since the index of the summation in our question starts from n=3, we could shift the index of this series to n=3. I tried that, but it didnt

Jeffrey Jordon

Expert2021-12-16Added 2605 answers

Answer is given below (on video)

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