What is the sum of $n$ terms in the series

Question

What is the sum of $n$ terms in the series : $$\log m+\log \left(\dfrac{m^2}{n}\right) +\log \left(\dfrac{m^4}{n^3}\right) +\cdots\cdots$$

Options

$ a.)\ \log \left[\dfrac{n^{(n-1)}}{m^{(n+1)}}\right]^{n/2} \\ b.)\ \log \left[\dfrac{m^{m}}{n^{n}}\right]^{n/2} \\ c.)\ \log \left[\dfrac{m^{(1-n)}}{n^{(1-m)}}\right]^{n/2} \\ \color{green}{d.)\ \log \left[\dfrac{m^{(n+1)}}{n^{(n-1)}}\right]^{n/2}} \\ $

I am highly confused on how to interpret this question , and it looks very ambiguous to me.

I tried the following

$ \log m+\log \left(\dfrac{m^{2}}{n}\right) +\log \left(\dfrac{m^{4}}{n^{3}} \right) +\cdots \\ $

$\implies \log \left(\dfrac{m\times m^{2} \times m^{4} \times \cdots }{n^{0}\times n^{1}\times n^{3}\times \cdots }\right) \\ $

$ \implies \log \left( \dfrac{m^{1+2+4+ \cdots}}{n^{0+1+3+ \cdots}}\right) \\ $

I would like to also know the option $d.)$ which is given as the answer by book is actually correct or wrong.

I look for a simple and short way.

I have studied maths up to $12th$ grade.

Answer 1

HINT:

You are on right track, though the number of terms supplied is too low to determine the actual pattern.

$1,2,4,\cdots$ seems to be a Geometric Series with common ratio $=2$

$1,3,\cdots$ seems to be a Arithmetic Series with common difference $=2$

Answer 2

At times working backward helps.

What does (d) represent?

Analyse the power of m and n in the answer (which = $n(n+1)/2$ and $n(n-1)/2$)

• m: differentiate the power once (with repect to to n) to get $(2n+1)/2$
• n: do the same to get $(2n-1)/2$

Now since the degree of n after differentiation is 1 in both the cases, I'm sure enough to assume the progressions of exponent(s) in your last step should be an arithmetic progression.

• for m: Let the first term be $a_m$ and constant difference be $d_m$. Therefore, I've $\frac {n} { 2} (2a + (n-1)d) = \frac {n/2} { n+1}$. Now compare the coefficient of n to get d and then the constant terms for a. Final values are: $a_m=1=d_m$.
• for n: repeat the same procedure to get $a_n=0$ and $d_n=1$ here.

Now that we've the series, a question with answer = (d) can be framed:

$log (\frac { m^{1+2+3+...}} {n^{0+1+2+3+...} })$

Therefore the correct "question" becomes: $log (m)+ log(\frac {m^2}{n}) + log (\frac{m^3}{n^2}) ...$

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And I've not mentioned the way to solve the above question since you already know that. Go ahead to get the answer.