PaperMath’s sum

Papermath’s sum states,

$\sum_{i=0}^{2n-1} {(x^2 \times 10^i)}=(\sum_{j=0}^{n-1}{(3x \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2x^2 \times 10^k)}$

Or

$x^2\sum_{i=0}^{2n-1} {10^i}=(x \sum_{j=0}^{n-1} {(3 \times 10^j)})^2 + x^2\sum_{k=0}^{n-1} {(2 \times 10^k)}$

For all real values of $x$ , this equation holds true for all nonnegative values of $n$ . When $x=1$ , this reduces to

$\sum_{i=0}^{2n-1} {10^i}=(\sum_{j=0}^{n -1}{(3 \times 10^j)})^2 + \sum_{k=0}^{n-1} {(2 \times 10^k)}$

Proof

First, note that the $x^2$ part is trivial multiplication, associativity, commutativity, and distributivity over addition,

Observing that $\sum_{i=0}^{n-1} {10^i} = (10^{n}-1)/9$ and $(10^{2n}-1)/9 = 9((10^{n}-1)/9)^2 + 2(10^n -1)/9$ concludes the proof.

Problems

AMC 12A Problem 25

For a positive integer $n$ and nonzero digits $a$ , $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$ ?