Find the sum to $n$ terms of the series $10+84+734+....$ Find the sum to n terms of the series $10+84+734+....$


*

*$\frac{9(9^n+1)}{10} + 1$

*$\frac{9(9^n-1)}{8} + 1  $

*$\frac{9(9^n-1)}{8} + n $

*$ \frac{9(9^n-1)}{8} + n^2$



My attempt:
I'm getting option $(4)$.
For $n=1$, all options are right,
For $n=2$, sum must be $10+84=94$.

Can you explain in formal way? Please.

 A: The terms are neither geometric nor arithmetic progression. There could be more than one recurrence relations that can describe the $n^{th}$ term, but it is unlikely that you could deduce it based on $3$ terms only. Since this is only a multiple choice questoon, I suspect that they expect you to find out the right choice using your method.
On the mathematical side, we can find out the general term $T(n)$ by subtracting the sum of first $n-1$ terms from the sum of first $n$ terms. (By the way I suppose the option is $\frac{9(9^n-1)}8+n^2$ 
$$T(n)=\frac{9^{n+1}-9}8 + n^2  - \frac{9^n-9}8 -(n-1)^2$$
$$T(n)=9^n +2n -1$$
Which is pretty hard to guess based on the first three terms only
A: REMARK.-If the first three terms $10+84+734+.....$ are correctly written then the formation law could be $a_n=9^n+2n-1$ However this conflicts with the four given options (in comment I suggested could be a typo).
A: Note: I think formal issues are not that essential. But you have to reasonably show that you have checked all variants and derive the appropriate conclusion. Here is one variation of the theme.

We have four options each providing a sequence 
  \begin{align*}
(a_n)_{n\geq 1}=(a_1,a_2,a_3,\ldots)
\end{align*}
The sequence against we have to check is
  \begin{align*}
\left(10,10+84,10+84+734,\ldots\right)=\left(10,94,828,\ldots\right)
\end{align*}

We do so by listing each of the four alternatives as far as necessary:
\begin{align*}
\left(\frac{9(9^n+1)}{10} + 1\right)_{n\geq 1}&=(10,74.1,\ldots)\tag{1}\\
\left(\frac{9(9^n-1)}{8} + 1\right)_{n\geq 1}&=(10,91,\ldots)\tag{2}\\
\left(\frac{9(9^n-1)}{8} + n\right)_{n\geq 1}&=(10,92,\ldots)\tag{3}\\
\left(\frac{9(9^n-1)}{8} + n^2\right)_{n\geq 1}&=(10,94,828,\ldots)\tag{4}\\
\end{align*}

We observe alternatives (1) to (3) are no proper solution, since the second element of each of these three alternatives is not equal 
  \begin{align*}
10+84=94
\end{align*}
  The last option (4) coincides in all three values $(10,10+84,10+84+734)=(10,94,828)$ with the series which is to generate.
and conclude the sequence
  \begin{align*}
\left(\frac{9(9^n-1)}{8} + n^2\right)_{n\geq 1}=(10,94,828,\ldots)
\end{align*}
  is the solution.

A: The answer is option 4.
Let $S(n)=\frac{9(9^n-1)}{8} + n^2$
Then $S(1)=10$, matching with the given series.
$S(2)=94$ which is equal to $10+84$(sum to $2$ terms), matching with the given series.
$S(3)=828=10+84+734$(sum to $3$ terms), again matching with the given series.
A: Although it is an old question, I'd like to contribute a remark of mine on this, especially, after my long absence from the site. I realized at one glance that the sequence is $$9 + 1, 9^2 + 3, 9^3 + 5, \cdots,$$ which led me to think that the general term of the sequence is $9^n + 2 n + 1$. Summing from $1$ to $n$, we have $$9 (9^n - 1) / 8 + n^2.$$
