# Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 9"

## Problem

Let $C_n = 1+10 +10^2 + \cdots + 10^{n-1}.$

(a) Prove that $9C_n = 10^n -1.$

(b) Prove that $(3C_3+ 2)^2 =112225.$

(c) Prove that each term in the following sequence is a perfect square: $$25, 1225, 112225, 11122225, 1111222225,\ldots$$

## Solution

(a) We know that $C_n$ is a geometric series, so we can define it explicitly as follows

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

multiplying both sides by 9 yields our answer.

(b) We have

$(3*111+2)^2=335^2$,

yielding $112,225$.

(c) We say that the nth member of the sequence equals $(3*C_n+2)^2$. Expanding yields

$(3*\frac{10^n-1}{9}+2)^2$,

$=(\frac{10^n+5}{3})^2$,

$=\frac{10^{2n}}{9}+\frac{10^{n+1}}{9}+\frac{25}{9}$.

Dividing each term separately, we know that the first term will add $2n$ $1$s and $\frac{1}{9}$, the second term will add $n+1$ $1$s and $\frac{1}{9}$, and the third will add $\frac{25}{9}$, giving

$\overbrace{11...11}^{2n}+\overbrace{11...11}^{n+1}+\frac{27}{9}$,

$\overbrace{11...11}^{n-1}\overbrace{22...22}^{n}5$,

which is exactly what we wanted.

(a) $\frac{10^n-1}{9}$ (b) $112,225$ (c) $11,122,225$