
The coefficient of \[{x^n}\] in the expansion of \[{\log _e}(1 + 3x + 2{x^2})\] is [UPSEAT\[2001\]]
A) \[{( - 1)^n}[\dfrac{{{2^n} + 1}}{n}]\]
B) \[\dfrac{{{{( - 1)}^{n + 1}}}}{n}[{2^n} + 1]\]
C) \[\dfrac{{{2^n} + 1}}{n}\]
D) None of these
Answer
164.1k+ views
Hint: in this question, we have to find the coefficient of given value\[{x^n}\]. In order to solve this first rearrange the given expression to find standard known expression. Once we get type of function then by applying the formula of that function, the required value is to be calculated.
Formula Used:\[\log (a \times b) = \log a + \log b\]
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = \sum\limits_{m = 1}^\infty {( - } {)^{m - 1}}\dfrac{{{x^n}}}{m}\]
Complete step by step solution:Given: \[{\log _e}(1 + 3x + 2{x^2})\]
Now we have to rearrange the above log function in order to find standard pattern of log function.
\[{\log _e}(1 + 3x + 2{x^2}) = {\log _e}(2{x^2} + 2x + x + 1)\]
\[{\log _e}(2{x^2} + 2x + x + 1) = {\log _e}(2x(1 + x) + (x + 1))\]
\[{\log _e}(2x(1 + x) + (x + 1)) = {\log _e}((1 + x)(2x + 1))\]
Now we know that
\[\log (a \times b) = \log a + \log b\]
\[{\log _e}((1 + x)(2x + 1)) = {\log _e}(1 + x) + {\log _e}(1 + 2x)\]
We also know that
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = \sum\limits_{m = 1}^\infty {( - } {)^{m - 1}}\dfrac{{{x^n}}}{m}\]
\[{\log _e}(1 + x) + {\log _e}(1 + 2x) = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{{x^n}}}{n}} + \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{2{x^n}}}{n}} \]
\[\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{{x^n}}}{n}} + \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{2{x^n}}}{n}} = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}(\dfrac{1}{n} + \dfrac{{{2^n}}}{n}){x^n}} \]
\[ = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}(\dfrac{{{2^n} + 1}}{n}){x^n}} \]
Coefficient of \[{x^n}\]\[ = \dfrac{{{{( - 1)}^{n + 1}}}}{n}[{2^n} + 1]\]
Required value is \[\dfrac{{{{( - 1)}^{n + 1}}}}{n}[{2^n} + 1]\]
Option ‘B’ is correct
Note: Here we have to rearrange the function in order to get standard pattern of function. Once we get type of function then by applying the formula of that series, the required value is to be calculated.
In this question after rearrangement we found that function is written in the form of\[{\log _e}(1 + x)\]. After getting standard series we have to apply the expansion formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.
Formula Used:\[\log (a \times b) = \log a + \log b\]
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = \sum\limits_{m = 1}^\infty {( - } {)^{m - 1}}\dfrac{{{x^n}}}{m}\]
Complete step by step solution:Given: \[{\log _e}(1 + 3x + 2{x^2})\]
Now we have to rearrange the above log function in order to find standard pattern of log function.
\[{\log _e}(1 + 3x + 2{x^2}) = {\log _e}(2{x^2} + 2x + x + 1)\]
\[{\log _e}(2{x^2} + 2x + x + 1) = {\log _e}(2x(1 + x) + (x + 1))\]
\[{\log _e}(2x(1 + x) + (x + 1)) = {\log _e}((1 + x)(2x + 1))\]
Now we know that
\[\log (a \times b) = \log a + \log b\]
\[{\log _e}((1 + x)(2x + 1)) = {\log _e}(1 + x) + {\log _e}(1 + 2x)\]
We also know that
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = \sum\limits_{m = 1}^\infty {( - } {)^{m - 1}}\dfrac{{{x^n}}}{m}\]
\[{\log _e}(1 + x) + {\log _e}(1 + 2x) = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{{x^n}}}{n}} + \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{2{x^n}}}{n}} \]
\[\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{{x^n}}}{n}} + \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\dfrac{{2{x^n}}}{n}} = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}(\dfrac{1}{n} + \dfrac{{{2^n}}}{n}){x^n}} \]
\[ = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}(\dfrac{{{2^n} + 1}}{n}){x^n}} \]
Coefficient of \[{x^n}\]\[ = \dfrac{{{{( - 1)}^{n + 1}}}}{n}[{2^n} + 1]\]
Required value is \[\dfrac{{{{( - 1)}^{n + 1}}}}{n}[{2^n} + 1]\]
Option ‘B’ is correct
Note: Here we have to rearrange the function in order to get standard pattern of function. Once we get type of function then by applying the formula of that series, the required value is to be calculated.
In this question after rearrangement we found that function is written in the form of\[{\log _e}(1 + x)\]. After getting standard series we have to apply the expansion formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
