How to find out the square of $1,111,111,111$ through induction

Question

I am aware of the answer being $1234567900987654321$. However I have been asked to prove it using mathematical induction. How should I go about it?

Answer 1

Noting that the sequence given can be expressed as $$ a_n=\underbrace{111 \cdots 1}_{n ’ 1’ s}=\sum_{k=0}^{n-1} 10^k=\frac{10^{n}-1}{9} $$ Therefore their squares are $$ \boxed{a_n^2=\left(\sum_{k=0}^{n-1} 10^k\right)^2=\left(\frac{10^{n}-1}{9}\right)^2=\frac{10^{2 n}-2 \cdot 10^{n}+1}{81}} $$ which can be also proved by induction.

For example, when $n=20,$ $$a_{20}^2=123456790123456790120987654320987654321$$

Answer 2

If you can just argue based on patterns, I second julio's suggestion of looking at $1^2, 11^2, 111^2$, identifying the pattern, and extrapolating to your case.

But if you have to use proper Principle of Mathematical Induction, then I suggest looking at it like this: $$11 = 1\times 10 + 1 \\ 111 = 11\times 10 + 1 \\ \vdots$$

So if you know the pattern holds for $n$ $1$'s, try using the above expansion to prove it holds for $n+1$ $1$'s