
The sum of the series \[\dfrac{2}{3} + \dfrac{8}{9} + \dfrac{{26}}{{27}} + \dfrac{{80}}{{81}} + \]……… n terms are:
A) \[n - \dfrac{1}{2}({3^n} - 1)\]
B) \[n + \dfrac{1}{2}({3^n} - 1)\]
C) \[n - \dfrac{1}{2}(1 - {3^{ - n}})\]
D) \[n + \dfrac{1}{2}({3^{ - n}} - 1)\]
Answer
163.5k+ views
Hint: in this question we have to find n term of given series. Here we have to first find the pattern of the series. Rearrange the given series to and check for pattern. If any patterns is observed then follow the same pattern and use appropriate formula to get the required value.
Formula Used: We can find sum of n terms of GP by using
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
Complete step by step solution: Given: \[\dfrac{2}{3} + \dfrac{8}{9} + \dfrac{{26}}{{27}} + \dfrac{{80}}{{81}} + \]
Now rearrange the above series
\[(1 - \dfrac{1}{3}) + (1 - \dfrac{1}{{{3^2}}}) + (1 - \dfrac{1}{{{3^3}}}) + ... + (1 - \dfrac{1}{{{3^n}}})\]
\[(1 + 1 + 1 + 1 + 1 + ..... + n) - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
\[(\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\] This series are in GP
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Here \[a = \dfrac{1}{3}\]and \[r = \dfrac{1}{3}\]
We know that Sum of n terms of GP is given by
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Now apply sum formula
\[ = n - \dfrac{{\dfrac{1}{3}\{ 1 - {{(\dfrac{1}{3})}^n}\} }}{{1 - (\dfrac{1}{3})}}\]
Now required value is
\[ = n + \dfrac{1}{2}({3^{ - n}} - 1)\]
Option ‘D’ is correct
Note: Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio therefore always try to find first term and common ratio.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Formula Used: We can find sum of n terms of GP by using
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
Complete step by step solution: Given: \[\dfrac{2}{3} + \dfrac{8}{9} + \dfrac{{26}}{{27}} + \dfrac{{80}}{{81}} + \]
Now rearrange the above series
\[(1 - \dfrac{1}{3}) + (1 - \dfrac{1}{{{3^2}}}) + (1 - \dfrac{1}{{{3^3}}}) + ... + (1 - \dfrac{1}{{{3^n}}})\]
\[(1 + 1 + 1 + 1 + 1 + ..... + n) - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
\[(\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\] This series are in GP
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Here \[a = \dfrac{1}{3}\]and \[r = \dfrac{1}{3}\]
We know that Sum of n terms of GP is given by
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Now apply sum formula
\[ = n - \dfrac{{\dfrac{1}{3}\{ 1 - {{(\dfrac{1}{3})}^n}\} }}{{1 - (\dfrac{1}{3})}}\]
Now required value is
\[ = n + \dfrac{1}{2}({3^{ - n}} - 1)\]
Option ‘D’ is correct
Note: Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio therefore always try to find first term and common ratio.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Recently Updated Pages
JEE Advanced Percentile vs Marks 2025| Previous year's trends

JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

Trending doubts
IIT Kanpur Highest Package, Average & Median Salary

IMU CET SYLLABUS 2025

Difference Between Line Voltage and Phase Voltage

IIT Indore Average Package: Placement Overview

JEE Advanced Syllabus 2025 (OUT)

IIT Hyderabad Highest Package 2025: Detailed Placement Insights

Other Pages
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government College Seat Matrix
