# a) Find the Maclaurin series for the function f(x)=frac11+x b) Use differentiation of power series and the result of part a) to find the Maclaurin ser

a) Find the Maclaurin series for the function
$$f(x)=\frac11+x$$
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
$$\sum_{n=3}^\infty \frac{n(n-1)}{2n}$$
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 $$\frac72$$

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Nathalie Redfern
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:
$$f(x)=\frac{1}{(1+x)^3}=\sum_{n=0}^\infty\frac12(-1)^n(n+1)(n+2)x^n$$
We can similarly write a series for $$\frac{1}{(1-x)^3}$$ as shown below:
$$f(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^\infty\frac12(n+1)(n+2)x^n$$
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 didn't work. What worked was shifting the index to n = 2. So, upon shifting the index to n=2, we get:
$$f(x)=\frac{1}{(1-x)^3}=\sum_{n=2}^\infty\frac12(n-1)nx^{n-2}$$
Since we need $$2^n$$ in the denominator. We can get that by substituting $$x=\frac12$$ in our previous step. Upon doing so, and simplify a little bit, we get:
$$f(\frac12)=\frac{1}{(1-\frac12)^3}=\sum_{n=2}^\infty\frac12(n-1)n(\frac12)^{n-2}$$
$$\Rightarrow\frac{1}{(\frac12)^3}=\sum_{n=2}^\infty\frac12(n-1)n\frac{1}{2^{n-2}}$$
$$\Rightarrow\frac{1}{\frac18}=\sum_{n=2}^\infty\frac{(n-1)n}{2^{n-1}}\Rightarrow8=\sum_{n=2}^\infty\frac{n(n-1)}{2^{n-1}}$$
We have $$2^{n-1}$$ in the denominator of right hand side. In order to get $$2^n$$, we will divide both the sides by 2 and get:
$$\Rightarrow4=\sum_{n=2}^\infty\frac{n(n-1)}{2^n}$$
Now we can work out the index. Our index starts from 2 and we need it to start from 3. So, we can split this series by finding the sum of the first term (corresponding to n=2) and the series from n=3 to infinity as shown below:
$$4=\frac{2(2-1)}{2^2}=\sum_{n=3}^\infty\frac{n(n-1)}{2^n}$$
Finally, we can simplify it and bring the constant numbers on one side to get the required answer:
$$\Rightarrow4=\frac{2(2-1)}{2^2}+\sum_{n=3}^\infty\frac{n(n-1)}{2^n}\Rightarrow4=\frac24+\sum_{n=3}^\infty\frac{n(n-1)}{2^n}$$
$$\Rightarrow4=\frac12+\sum_{n=3}^\infty\frac{n(n-1)}{2^n}\Rightarrow4-\frac12=\sum_{n=3}^\infty\frac{n(n-1)}{2^n}$$
$$\Rightarrow\frac72=\sum_{n=3}^\infty\frac{n(n-1)}{2^n}$$
Answer: $$\frac72$$
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