Later data often show patterns better than early data, so extend your table of partial sums a bit:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171
\end{array}$$
Ignoring the signs, it appears that the numbers in the bottom line are approximately doubling each time. Moreover, still ignoring signs, adjacent partial sums add up to a power of $2$: $|s_1|+|s_2|=2^1$, $|s_2|+|s_3|=2^2$, $|s_3|+|s_4|=2^3$, and apparently in general $|s_n|+|s_{n+1}|=2^n$. (If you go back to the definition of the partial sums, you’ll see why this happens.)
If $|s_n|+|s_{n+1}|=2^n$ and $|s_{n+1}|\approx 2|s_n|$, then $3|s_n|\approx 2^n$; this suggests that we should compare $3|s_n|$ with $2^n$:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9\\
s_n:&1&-1&3&-5&11&-21&43&-85&171\\
3|s_n|:&3&3&9&15&33&63&129&255&513\\
2^n:&2&4&8&16&32&64&128&256&512
\end{array}$$
That pattern’s pretty clear: apparently $3|s_n|=2^n+1$ if $n$ is odd, and $3|s_n|=2^n-1$ if $n$ is even. Those cases can be combined as $3|s_n|=2^n+(-1)^{n+1}$, since $(-1)^{n+1}$ is $1$ when $n$ is odd and $-1$ when $n$ is even. And the algebraic sign of $s_n$ appears to be that of $(-1)^{n+1}$, so if these patterns are real,
$$\begin{align*}
s_n&=\frac{(-1)^{n+1}}3\left(2^n+(-1)^{n+1}\right)\\
&=\frac{(-1)^{n+1}2^n}3+\frac{(-1)^{2n+2}}3\\
&=\frac{(-1)^{n+1}2^n+1}3\\
&=\frac{1-(-2)^n}3\;.
\end{align*}$$
This result can then be proved by mathematical induction, but I suspect that you’re not expected to go that far.
If you’ve already learned the summation formula for finite geometric series, you can apply it to get $s_n$ without looking at any patterns at all, and it’s something that you should learn as soon as possible if you don’t already know it. However, skill at pattern-recognition is useful anyway, so I thought that it might be useful to see how the problem can be attacked in that way as well.
Yes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I mean? Consider this
$$
t_n = n+n^2
$$
Subsituiting $n=1,2,3,\cdots$ generates your terms. By the way, $202$ doesn't occur in this sequence, the 13th term is $182$ and the $14th$ term is $210$. I am assuming it was supposed to be $210$.
So we need to find
$$
\sum_{i=1}^{n}i+i^2 = \sum_{i=1}^{n}i+\sum_{i=1}^{n}i^2
$$
where $n=14$. There are well known formulas for $\sum_{i=1}^{n}i$ and for $\sum_{i=1}^{n}i^2$. Substituting them, we get,
$$\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$$
$$=\frac{n(n+1)}{2}\left(1+\frac{2n+1}{3}\right)$$
$$=\frac{n(n+1)}{2}\left(\frac{3+2n+1}{3}\right)$$
$$=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)$$
$$=\frac{n(n+1)(n+2)}{3}$$
where $n=14$. Thus our sum is $1120$.
Best Answer
It’s the formula for the sum of a finite geometric series:
$$\sum_{k=0}^{n-1}x^k=\frac{1-x^n}{1-x}\;,$$
here with $x=\frac13$. Specifically,
$$\sum_{k=1}^n\frac1{3^{k-1}}=\sum_{k=0}^{n-1}\frac1{3^k}=\sum_{k=0}^{n-1}\left(\frac13\right)^k=\frac{1-(1/3)^n}{1-1/3}=\frac32\left(1-\left(\frac13\right)^n\right)=\frac32\left(1-\frac1{3^n}\right)\;.$$