<s>[INST] 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.

 [/INST] 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.

</s>