This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$\frac{4n(n+1)(2n+1)}{6} + n + \frac{4(n\cdot n+1)}{2}$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
Best Answer
A simple answer to the question
Note that \begin{multline*} S\equiv\sum_{n\geq1}\frac{(2n-1)^{2}}{2^{n}}=\sum_{n\geq1}\frac{4n^{2}-4n+1}{2^{n}}=4\sum_{n\geq1}\frac{n^{2}}{2^{n}}-4\sum_{n\geq1}\frac{n}{2^{n}}+\sum_{n\geq1}\frac{1}{2^{n}}\equiv4I_2-4I_1+I_0. \end{multline*} since each of the series on the right hand side are convergent. First, note that $I_0$ is a geometric series with $I_0=1$ (you can find an exposition to geometric series on Wikipedia). As for $I_1$, \begin{align*} I_1 & =\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots\\ \frac{1}{2}I_1 & =\frac{0}{2^{1}}+\frac{1}{2^{2}}+\frac{2}{2^{3}}+\cdots\\ \frac{1}{2}I_1=I_1-\frac{1}{2}I_1 & =\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots \end{align*} and hence $\frac{1}{2}I_1$ is once again a geometric series with $\frac{1}{2}I_1=1$ so that $I_1=2$. As for $I_2$, note that \begin{align*} I_2 & =\frac{1}{2^{1}}+\frac{4}{2^{2}}+\frac{9}{2^{3}}+\cdots\\ \frac{1}{2}I_2 & =\frac{0}{2^{1}}+\frac{1}{2^{2}}+\frac{4}{2^{3}}+\cdots\\ \frac{1}{2}I_2=I_2-\frac{1}{2}I_2 & =\frac{1}{2^{1}}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots \end{align*} In other words, $$ \frac{1}{2}I_2=\sum_{n\geq1}\frac{2n-1}{2^{n}}=2I_1-I_0=2\cdot2-1=3. $$ Putting this all together, $$ S=4\cdot6-4\cdot2+1=24-8+1=17. $$
Generalizing the above
We can generalize the approach above. Fix a constant $c\in\mathbb{C}$ with $|c|>1$. For each nonnegative integer $m$, let $$ \boxed{I_{m}=\sum_{n\geq1}n^{m}c^{-n}} $$ Note that $I_{0}=(c-1)^{-1}$. If $m>0$, then \begin{multline*} \left(c-1\right)c^{-1}I_m=I_m-c^{-1}I_m=\sum_{n\geq1}n^{m}c^{-n}-\sum_{n\geq1}n^{m}c^{-n-1}\\ =\sum_{n\geq1}n^{m}c^{-n}-\sum_{n\geq1}\left(n-1\right)^{m}c^{-n}=\sum_{n\geq1}\left(n^{m}-\left(n-1\right)^{m}\right)c^{-n}\\ =\sum_{n\geq1}\left(n^{m}-\sum_{k=0}^{m}\binom{m}{k}n^{k}\left(-1\right)^{m-k}\right)c^{-n} =\sum_{n\geq1}\left(\sum_{k=0}^{m-1}\binom{m}{k}n^{k}\left(-1\right)^{m-1-k}\right)c^{-n}\\ =\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}I_{k}. \end{multline*} This yields the recurrence $$ \boxed{I_{m}=\left(c-1\right)^{-1}c\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}I_{k}\qquad\text{for }m\geq 1} $$
Relationship to polylogarithm
Note that $I_{m}$ is related to the polylogarithm: $$ \operatorname{Li}_{-m}(c^{-1})=\sum_{n\geq1}\frac{\left(c^{-1}\right)^{n}}{n^{-m}}=\sum_{n\geq1}\frac{c^{-n}}{n^{-m}}=\sum_{n\geq1}n^{m}c^{-n}=I_{m}. $$ Expressing our identity in terms of the polylogarithm, $$ \boxed{\operatorname{Li}_{-m}(z)=\left(1-z\right)^{-1}\sum_{k=0}^{m-1}\binom{m}{k}\left(-1\right)^{m-1-k}\operatorname{Li}_{-k}(z)\qquad\text{for }m\geq 1\text{ and }0<|z|<1} $$ There are various other expressions for $\operatorname{Li}_{-m}(z)$ where $m$ is positive and $z$ is complex.