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Question: The coefficient of \(x^{n}\) in the expansion of \(\log_{e}(1 + 3x + 2x^{2})\) is...

The coefficient of xnx^{n} in the expansion of

log⁡e(1+3x+2x2)\log_{e}(1 + 3x + 2x^{2}) is

A

(−1)n[2n+1n]( - 1)^{n}\left\lbrack \frac{2^{n} + 1}{n} \right\rbrack

B

(−1)n+1n[2n+1]\frac{( - 1)^{n + 1}}{n}\lbrack 2^{n} + 1\rbrack

C

2n+1n\frac{2^{n} + 1}{n}

D

None of these

Answer

(−1)n+1n[2n+1]\frac{( - 1)^{n + 1}}{n}\lbrack 2^{n} + 1\rbrack

Explanation

Solution

We have, log⁡e(1+3x+2x2)=log⁡e(1+x)+log⁡e(1+2x)\log_{e}(1 + 3x + 2x^{2}) = \log_{e}(1 + x) + \log_{e}(1 + 2x)

=∑n=1∞(−1)n−1xnn+∑n=1∞(−1)n−1(2x)nn=∑n=1∞(−1)n−1(1n+2nn)xn=∑n=1∞(−1)n−1(1+2nn)xn= \sum_{n = 1}^{\infty}{( - 1)^{n - 1}\frac{x^{n}}{n} + \sum_{n = 1}^{\infty}{( - 1)^{n - 1}\frac{(2x)^{n}}{n}}} = \sum_{n = 1}^{\infty}{( - 1)^{n - 1}\left( \frac{1}{n} + \frac{2^{n}}{n} \right)x^{n} = \sum_{n = 1}^{\infty}{( - 1)^{n - 1}}\left( \frac{1 + 2^{n}}{n} \right)x^{n}}

So coefficient of xn=(−1)n−1(2n+1n)=(−1)n+1(2n+1)nx^{n} = ( - 1)^{n - 1}\left( \frac{2^{n} + 1}{n} \right) = \frac{( - 1)^{n + 1}(2^{n} + 1)}{n}

[∵(−1)n=(−1)n+2=........]\left\lbrack \because( - 1)^{n} = ( - 1)^{n + 2} = ........ \right\rbrack