2008 UNCO Math Contest II Problems/Problem 9


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: \[25, 1225, 112225, 11122225, 1111222225,\ldots\]


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


multiplying both sides by 9 yields our answer.

(b) We have


yielding $112,225$.

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




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



which is exactly what we wanted.

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

Solution 2

(a) $9=10-1$. Multiply these two binomials and we have reach our answer (remember the formula -- it's like Difference of Cubes)

(b)$C_3=111$. The original expression is equal to $\boxed{112225}$. (Just brute force this out).

(c) Now notice that each term in the sequence is $(3C_n)^2$. As seen in part (a), we see that $C_n=\frac{10^n-1}{9}$. Follow Solution 1 above.


See Also

2008 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions