Question

$(\mathrm{666...}n$ digits${)}^2+(\mathrm{888...}n$ digits${)}^2$

• A. $\frac{4}{9}({10}^n-1)$
• B. $\frac{4}{9}({10}^{2n}-1)$
• C. $\frac{4}{9}{({10}^n-1)}^2$
• D. $\frac{100}{81}{({10}^n-1)}^2$

Answer 1

The correct option is D $\frac{100}{81}{({10}^n-1)}^2$

$=6^2\times (\mathrm{11111...}n{)}^2+8^2\times (\mathrm{11111...}n{)}^2$
$=100(\mathrm{1111...}n{)}^2$
$=100[{10}^{n-1}+{10}^{n-2}....1{]}^2$(Number can be written in the following form)
So this is a GP with $a=1r=10$terms=$n$
$=100\times [\frac{{10}^n-1}{9}{]}^2$
So the correct answer is $100\times \frac{({10}^n-1{)}^2}{81}$
So D is the correct answer