Question

The sum to $n$ terms of the series $2+5+14+41+....$ is
A. \[{{3}^{n-1}}+8n-3\]
B. \[{{8.3}^{n}}+4n-8\]
C. \[{{3}^{n+1}}+\dfrac{8}{3}n+1\]
D. None of these

Answer 1

Hint: In this question, we are to find the sum to $n$ terms of the given series. To do this, we first need to know the type of the series. Here by simplifying and rewriting the series, we get the type of progression. Then, by applying appropriate formulae, we can calculate the required sum.

Formula used: If the series is an Arithmetic series, then the sum of the $n$ terms of the series is calculated by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
Where the common difference $d={{a}_{n}}-{{a}_{n-1}}$
If the series is a geometric series, then the sum of the $n$ terms of the series is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$
Where the common ratio $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum to the $n$ terms of the series; $a$ - The first term; $n$ - Number of terms of the series.

Complete step by step solution: Given that the series is
${{S}_{n}}=2+5+14+41+....\text{ }(1)$
On rewriting the series, we get
$\begin{align}
  & \Rightarrow {{S}_{n}}=2+(2+3)+(2+3+9)+(2+3+9+27)+.... \\
 & \Rightarrow {{S}_{n}}=2+(2+3)+(2+3+{{3}^{2}})+(2+3+{{3}^{2}}+{{3}^{3}})+...\text{ }(2) \\
\end{align}$
Then, we can write the $nth$ term of the series (2) is
${{x}_{n}}=2+\left( 3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{n}} \right)$
Then, we can evaluate the $n^{th}$ term of the series where the inner series is a geometric series, we can find its sum. I.e.,
$\begin{align}
  & {{x}_{n}}=2+\left( 3+{{3}^{2}}+{{3}^{3}}+...+{{3}^{n-1}} \right) \\
 & \text{ }=2+\left( \dfrac{3({{3}^{n-1}}-1)}{3-1} \right) \\
 & \text{ }=2+\dfrac{3}{2}({{3}^{n-1}}-1) \\
 & \text{ }=2+\dfrac{{{3}^{n}}}{2}-\dfrac{3}{2} \\
 & \text{ }=\dfrac{1}{2}+\dfrac{{{3}^{n}}}{2} \\
\end{align}$
Then, the series at (2) becomes
\[\begin{align}
  & {{S}_{n}}=2+(2+3)+(2+3+{{3}^{2}})+(2+3+{{3}^{2}}+{{3}^{3}})+...{{x}_{n}} \\
 & \text{ }={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \\
 & \text{ }=\sum{{{x}_{n}}} \\
 & \text{ }=\sum{\left( \dfrac{1}{2}+\dfrac{1}{2}{{3}^{n}} \right)} \\
\end{align}\]
On simplifying, we get
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{1}{2}\sum{n}+\dfrac{1}{2}\sum{{{3}^{n}}} \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}+\dfrac{1}{2}\left( \dfrac{3\left( {{3}^{n}}-1 \right)}{3-1} \right) \\
 & \therefore {{S}_{n}}=\dfrac{n}{2}+\dfrac{3}{4}\left( {{3}^{n}}-1 \right) \\
\end{align}$

Thus, Option (D) is correct.

Note: Here we need to rewrite the given series in order to find the required sum to the $n$ terms of the series. The inner series is in the geometric progression. So, by using the sum to $n$ formula, we can calculate the required sum.