
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
161.7k+ views
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.
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
