Question

Find the sum of the series \[2 + 5 + 14 + 41 + 122 + ....\] up to n terms and hence evaluate \[{S_8}\].

Answer 1

Hint: A series is a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. The sum of a series is the sum of the terms of the given series. An infinite series has infinite numbers of terms. The sum \[{S_n}\] of the first n terms of an infinite series is called a partial sum.
Some of the most common types of series are Arithmetic series, Geometric series, Harmonic series, and Fibonacci Numbers series.
First, find the nature of the progression of the given series. The behavior of the series depends on the common difference and the ratio; if it is positive then, the sequence is increasing towards the infinity, and if the difference is negative, then the series is decreasing to the negative infinity.

Complete step by step answer:
The given series is \[2 + 5 + 14 + 41 + 122 + ....\]
The differences of the consecutive term of the series are in G.P
\[3,9,27,81,......\]
For the sum of the given series, let the \[{n^{th}}\] term of the series be \[{t_n}\]; hence it can be written as
\[{S_n} = 2 + 5 + 14 + 41 + 122 + .... + {t_{n - 1}} + {t_n} - - - (i)\]
By increasing one place, this series can be written as
\[{S_n} = \mathop \_\nolimits^{} 2 + 5 + 14 + 41 + 122 + .... + {t_{n - 1}} + {t_n} - - - (ii)\]
Now subtract equation (ii) from (i), we get:
\[
{S_n} = 2 + 5 + 14 + 41 + 122 + .... + {t_{n - 1}} + {t_n} \\
\underline {{S_n} = {}_ - \mathop {}\nolimits^{} 2 + 5 + 14 + 41 + 122 + .... + {t_{n - 1}} + {t_n}} \\
0 = 2 + 3 + 9 + 27 + 81........ + {3^{n - 1}} + \left( { - {t_n}} \right) \\
\]
Here, ${3^{n - 1}}$ is the second-last term as we can see that the terms are increasing as the raised power of 3 only. So, the term can be written as:
\[
0 = 2 + 3 + 9 + 27 + 81........ + {3^{n - 1}} + \left( { - {t_n}} \right) \\
0 = 2 + 3 + {3^2} + {3^3} + {3^4}........ + {3^{n - 1}} + \left( { - {t_n}} \right) \\
\]
Since in the sequence after the first term, the sequence is in G.P, hence it can be written as
\[
0 = 2 + \left( {3 + {3^2} + {3^3} + {3^4}........ + {3^{n - 1}}} \right) + \left( { - {t_n}} \right) \\
{t_n} = 2 + \dfrac{{3\left( {{3^{n - 1}} - 1} \right)}}{{3 - 1}} \\
= 2 + \dfrac{{3\left( {{3^{n - 1}} - 1} \right)}}{2} \\
= \dfrac{{4 + {3^n} - 3}}{2} \\
= \dfrac{{{3^n} + 1}}{2} \\
\]
Hence the \[{n^{th}}\]of the sequence is \[{t_n} = \dfrac{{{3^n} + 1}}{2}\]which satisfies the sequence,
Now for the sum of the sequence till \[{t_n}\] can be written as
\[
{S_n} = \sum\limits_{n = 1}^n {\dfrac{{{3^n} + 1}}{2}} \\
= \dfrac{1}{2}\sum\limits_{n = 1}^n {{3^{^n}} + 1} \\
\]
Hence by substituting the values of n=0, 1, 2, 3, 4… in the sum
\[
{S_n} = \dfrac{1}{2}\sum\limits_{n = 1}^n {{3^{^n}} + 1} \\
= \dfrac{1}{2}\left[ {\left( {{3^1} + {3^2} + {3^3} + ..... + {3^n}} \right) + n} \right] \\
\]
The submission of a G.P. series is \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}};r > 1\], hence
\[
{S_n} = \dfrac{1}{2}\left[ {\left( {{3^1} + {3^2} + {3^3} + ..... + {3^n}} \right) + n} \right] \\
= \dfrac{1}{2}\left[ {\dfrac{{3\left( {{3^n} - 1} \right)}}{2} + n} \right] \\
\]
Hence the sum of the series\[2 + 5 + 14 + 41 + 122 + ....\]up to n terms is
\[{S_n} = \dfrac{1}{2}\left[ {\dfrac{{3\left( {{3^n} - 1} \right)}}{2} + n} \right]\]
Now for the\[{S_8}\], we get:
\[
{S_n} = \dfrac{1}{2}\left[ {\dfrac{{3\left( {{3^n} - 1} \right)}}{2} + n} \right] \\
{S_8} = \dfrac{1}{2}\left[ {\dfrac{{3\left( {{3^8} - 1} \right)}}{2} + 8} \right] \\
\]

Note: Before finding the sum of the series, it is necessary to study the nature of the series since it makes the process easy and reduces the number of steps to find the answer. Students often make mistakes while approximating the type of series and applying the formula for the sum.