# 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

• Solve for $x$ does not seem like the right question, since $x$ is a dummy variable of summation. Maybe you want the sum in terms of $n$. We can split into two parts, an arithmetic series and a finite geometric series. There are simple formulas for each part. Commented Jan 9, 2015 at 3:47
• What do you want to solve for $x$ ? What is the equation to solve ? And, as André Nicolas pointed out, is $x$ the index for summation ? Please clarify. Commented Jan 9, 2015 at 3:51
• @AndréNicolas You're right, I worded it incorrectly. I'm not very well-versed in math, so I apologize. Commented Jan 9, 2015 at 3:52
• This is just nonsense, I do not think this is a real question. Commented Jan 9, 2015 at 3:52
• @VividD: After a while, what is intended will become clear. Commented Jan 9, 2015 at 3:56

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.