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: \(\frac72\)

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: \(\frac72\)