Sum of a series given the formula for the nth term So I have a series $1+1+3+9+23+\cdots$ and the formula given to find the $n^{th}$ term $R_n=1-2n+2^n$. The first part is to verify that the term $115$ exists in this series. I knew that if I just solve for $n$ in $1-2n+2^n=115$ and if $n$ is an integer solution then the term $115$ exists. In fact I could't solve it and I just guessed and checked that $n=7$. So can someone guide me to solve the following equation?
\begin{align}
1-2n+2^n &amp;= 115\\
-2n+2^n &amp;= 114
\end{align}
Second part of the question is, how can I find the sum of the first 25 terms of this series?
Since

$$ 1-2\cdot n+2^n\ =\ 115 $$
then
$$ 2^n &gt; 115 $$

hence

$$ n &gt; \log_2(115) &gt; 6 $$

Also, by induction,

$$ \forall_{n\ge 4}\quad 2^{n-1} &gt; 2\cdot n-1 $$

Thus,
$$ \forall_{n\ge 4}\quad 2^{n-1}\ &lt; 2^n-2\cdot n + 1
       \ =\ 115 $$

hence (for $\ n\ge 4$)
$$ n-1\ &lt;\ \log_2 115 &lt; 7 $$
i.e.
$$ n\ &lt;\ 8 $$

It follows that IF there is any integer solution $n$
it has t be $\ n=7.\ $ But is there any? The calculation

$$ 2^7-2\cdot 7 + 1\ =\ 115 $$

shows that indeed, $\ n:=7\ $ is a solution, and it is the ONLY integer solution.

REMARK But is there any other REAL solution? -- it would still have to satisfy inequality

$$ 6 &lt; n &lt; 8 $$

But looking at the derivative of $\ 2^x-2\cdot x-1,\ $
which is strictly positive when $\ x&gt;6,\ $ we see that there is no real solution $\ n\ $ but $\ n=7.$
Your guess and check is a good approach when you know $n$ must be an integer.  $2^n$ changes quickly compared with $2n$, so just ignore the $2n$ and find the $n$ which is close.  You can prove that $2^n-2n$ is increasing with $n$ for $n \gt 2$, so if you miss you can just increment $n$ by $1$ and have a proof that there is no $n$.  

The $1-2n$ part is an arithmetic series and the $2^n$ part is geometric.  Sum each and add them together.