$1 + \frac 64 + \frac {14}8 + \frac {30}{16} + \ldots$ I have to calculate the sum upto n terms of the following:
$$1 + \frac 64 + \frac {14}8 + \frac {30}{16} + \ldots$$
I found the general term as:
$$T_n = \frac {an^2+bn+c}{2^{n}},$$ where a, b and c are determined as follows:
$$a+b+c=2; 4a+2b+c=6; 9a+3b+c=14.$$
Solving and applying:
$$T_n = \frac {2n^2-2n+2}{2^{n}} = \frac {n^2-n+1}{2^{n-1}}.$$
Now, how to compute:
$$\sum_{k=1}^{n} \frac {k^2-k+1}{2^{k-1}}$$

How about choosing $$T_n=\frac{\frac23n^3-2n^2+\frac{16}3n-2}{2^n}$$ whose elements match the few elements you mentioned and $$\sum_{n=1}^{\infty}T_n=14.$$
Here is an interesting discussion.
