Question

If $n$ is an integer then ${\mathrm{111...1}}_{2n-times}-{\mathrm{222...2}}_{n-times}$ is a perfect square of _____.

• A. $\frac{1}{2}\left({10}^n-1\right)$
• B. $\frac{1}{2}\left({10}^n+1\right)$
• C. $\frac{1}{3}\left({10}^n+1\right)$
• D. $\frac{1}{3}\left({10}^n-1\right)$

Answer 1

The correct option is D $\frac{1}{3}\left({10}^n-1\right)$

Given ${\mathrm{111...1}}_{2n-times}-{\mathrm{222...2}}_{n-times}$

$=\frac{1}{9}{\mathrm{999...9}}_{2n-times}-\frac{2}{9}{\mathrm{999...9}}_{n-times}$

$=\frac{1}{9}({10}^{2n}-1)-\frac{2}{9}({10}^n-1)$

$=\frac{1}{9}({10}^n+1)({10}^n-1)-\frac{2}{9}({10}^n-1)$

$=\frac{1}{9}({10}^n-1)({10}^n+1-2)$

$=\frac{1}{9}({10}^n-1)({10}^n-1)$

$\therefore N=\frac{1}{9}({10}^n-1{)}^2$

$\therefore \sqrt{N}=\frac{1}{3}({10}^n-1)$

square root of $n$ is $\frac{1}{3}({10}^n-1)$