Question

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 1

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.