Show that 2n "1" digits subtract n "2" digits is a perfect square.
I came across a difficult question in my IB HL Math book while revising for my Sequences and Series test that I wanted to share. I can't seem to figure it out.
Here's how I approached it so far:
- I found a series to generate 2n "1" digits and n "2" digits respectively: $$ \sum_{k=1}^{2n}10^{k-1} $$ $$ \sum_{k=1}^{n}2\cdot10^{k-1} $$
Then, I used the sum formula:
$$ S_{n} = \frac{U_{1}(1-r^{n})}{1 - r} $$ For generating n digits of 1: $$ S_{2n} = \frac{1(1-10^{2n})}{-9} $$ For generating n digits of 2: $$ S_{n} = \frac{2(1-10^{n})}{-9} $$
At this point, the question asks you to show that, when you subtract these terms, you are left with a perfect square.
If you do subtract these algebraically, you are left with: $$ \frac{1+10^{2n} - 2\cdot10^{n}}{9} $$
What I thought of doing was attempting to get the square root of this. The denominator leaves 3, but there doesn't seem to be a way to simplify the square root of the numerator expression and show that it's an integer.
Can somebody help me on a method to approach this?
Thanks!
Hint: You are almost done. Observe that $$1 + 10^{2n} - 2\cdot10^n = (10^n - 1)^2.$$
I did the following : $$ \eqalign{ & \sum_{k=1}^{2n}10^{k-1} - \sum_{k=1}^{n}2\cdot 10^{k-1} = \cr &= \sum_{k=1}^{n}10^{k-1} + \sum_{k=1}^{n}10^{n+k-1} - \sum_{k=1}^{n}10^{k-1} - \sum_{k=1}^{n}10^{k-1} \cr &= \sum_{k=1}^{n}10^{n+k-1} - \sum_{k=1}^{n}10^{k-1} \cr &= 10^n \cdot \sum_{k=1}^{n}10^{k-1} - \sum_{k=1}^{n}10^{k-1} \cr &= (10^n-1) \cdot \sum_{k=1}^{n}10^{k-1} \cr &= (10^n-1) \cdot {10^n -1 \over 10 -1} = {(10^n-1)^2 \over 3^2 } = \biggl({10^n -1 \over 3}\biggr)^2 } $$