Question

The expression $\frac{{\int }_0^n\left[x\right]dx}{{\int }_0^n\left\{x\right\}dx}$ where [x] and {x} are integral and fractional part of x and $nϵN$ is equal to

• A. $\frac{1}{n-1}$
• B. $n-1$
• C. ${\sum }_{r=1}^{n-1}r-{\sum }_{r=1}^{n-2}r$
• D. $\frac{{\prod }_{r=1}^{n}r-{\prod }_{r=2}^{n}r-1}{{\prod }_{r=2}^{n}r-1}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B

$n-1$

C

${\sum }_{r=1}^{n-1}r-{\sum }_{r=1}^{n-2}r$

D

$\frac{{\prod }_{r=1}^{n}r-{\prod }_{r=2}^{n}r-1}{{\prod }_{r=2}^{n}r-1}$

Expression $\frac{{\int }_0^n\left[x\right]dx}{{\int }_0^n\left\{x\right\}dx}$

$=\frac{{\int }_0^1\left[x\right]dx+{\int }_1^2\left[x\right]dx+...+{\int }_{n-1}^n\left[x\right]dx}{n{\int }_0^1\left\{x\right\}dx}$

$=\frac{0+1+2+...+(n-1)}{n\times \frac{1}{2}}=(n-1)$

Also,

${\sum }_{r=1}^{n-1}r-{\sum }_{r=1}^{n-2}r=(1+2+3+...+(n-1\left)\right)-(1+2+...+(n-2\left)\right)$

$\Rightarrow (n-1)$

and

$\frac{{\prod }_{r=1}^{n}r}{{\prod }_{r=2}^{n}r-1}-1\Rightarrow \frac{n\times (n-1)\times ...3\times 2\times 1}{(n-1)(n-2)....\times 3\times 2\times 1}-1$

$=n-1$