<s>[INST] Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation 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 = $17$
 [/INST] The general term is $\frac{(2k-1)^2}{2^k}$. Using that, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{(2k-1)^2}{2^k}
&=\sum_{k=1}^\infty\frac{4k(k-1)+1}{2^k}\tag1\\
&=\sum_{k=1}^\infty8\binom{k}{k-2}\left(\frac12\right)^k+\sum_{k=1}^\infty\binom{k-1}{k-1}\left(\frac12\right)^k\tag2\\
&=\sum_{k=1}^\infty8\binom{-3}{k-2}\left(-\frac12\right)^k-\sum_{k=1}^\infty\binom{-1}{k-1}\left(-\frac12\right)^k\tag3\\
&=2\left(1-\frac12\right)^{-3}+\frac12\left(1-\frac12\right)^{-1}\tag4\\[12pt]
&=17\tag5
\end{align}
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8\binom{k}{2}=8\binom{k}{k-2}$ for $k\ge2$ and $\binom{k-1}{k-1}=1$ for $k\ge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
</s>