<s>[INST] Summation to Equation I have a summation and I want to be able to find the sum for given $n$ without having to go through $1,\dots,n$.
$$\sum_{x=1}^{n - 1}x+300\cdot2^{x/7}$$
It's been awhile since I've done summations and I can't figure this one out. Also, this isn't for homework if that makes a difference.
Thanks
 [/INST] Might as well give an answer.  Note that we can split this sum into the following two sums: $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \sum_{x=1}^{n-1} x + \sum_{x=1}^{n-1} 300 \cdot 2^{x/7}.$$
Given that $$\sum_{x=1}^n x = \frac{n(n+1)}{2}$$ we know that $$\sum_{x=1}^{n-1} x = \frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$$  We're halfway done!  Turning our attention to the second sum, we have that
\begin{align*}
\sum_{x=1}^{n-1} 300 \cdot 2^{x/7} & = 300 \sum_{x=1}^{n-1} 2^{x/7} \\
& = 300 \left(\frac{2^{n/7} - 2^{1/7}}{2^{1/7} - 1}\right)
\end{align*}
by the formula for a finite geometric series.
Therefore, the total sum is $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \frac{n(n-1)}{2} + 300\left(\frac{2^{n/7} - \sqrt[7]{2}}{\sqrt[7]{2} - 1}\right).$$  Plug in $n$ and you will have your solution.
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