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) $11,122,225$

Solution 2

(a) $9=10-1$ . Multiply these two binomials and we have reach our answer (remember the formula!)

(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.

~hastapasta

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2008 UNCO Math Contest II Problems/Problem 9

Contents

Problem

Solution

Solution 2

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