Question

The value of $\log x+\log (1+\frac{1}{x})+\log (1+\frac{1}{1+x})+\log (1+\frac{1}{2+x})+\dots \cdots +\log (1+\frac{1}{(n-1+x)})$

• A. $\log \frac{x}{n}$
• B. $\log nx$
• C. $\log (n+x)$
• D. $\log (n-1)x$

Answer 1

The correct option is C $\log (n+x)$

$\log x+\log (1+\frac{1}{x})+\log (1+\frac{1}{1+x})+\log (1+\frac{1}{2+x})+\dots \dots \dots +\log (1+\frac{1}{(n-1+x)})$

Multiplying first $2$ terms and using property $\log n+\log m=\log \left(mn\right)$:

$=\log (x+1)+\log (1+\frac{1}{1+x})+\log (1+\frac{1}{2+x})+\dots \dots \dots +\log (1+\frac{1}{(n-1+x)})$

Multiplying first $2$ terms again:

$=\log (x+2)+\log (1+\frac{1}{2+x})+\dots \dots \dots +\log (1+\frac{1}{(n-1+x)})$

Continuing this series we get:

$=\log (n-1+x)+\log (1+\frac{1}{(n-1+x)})$

$=\log (n+x)$