Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 9"
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(a) <math>\frac{10^n-1}{9}</math> (b) <math>112,225</math> (c) <math>11,122,225</math> | (a) <math>\frac{10^n-1}{9}</math> (b) <math>112,225</math> (c) <math>11,122,225</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | (a) <math>9=10-1</math>. Multiply these two binomials and we have reach our answer (remember the formula!) | ||
+ | |||
+ | (b)<math>C_3=111</math>. The original expression is equal to <math>\boxed{112225}</math>. (Just brute force this out). | ||
+ | |||
+ | (c) Now notice that each term in the sequence is <math>(3C_n)^2</math>. As seen in part (a), we see that <math>C_n=\frac{10^n-1}{9}</math>. Follow Solution 1 above. | ||
+ | |||
+ | ~hastapasta | ||
== See Also == | == See Also == |
Revision as of 11:29, 27 April 2022
Contents
Problem
Let
(a) Prove that
(b) Prove that
(c) Prove that each term in the following sequence is a perfect square:
Solution
(a) We know that is a geometric series, so we can define it explicitly as follows
multiplying both sides by 9 yields our answer.
(b) We have
,
yielding .
(c) We say that the nth member of the sequence equals . Expanding yields
,
,
.
Dividing each term separately, we know that the first term will add s and , the second term will add s and , and the third will add , giving
,
,
which is exactly what we wanted.
(a) (b) (c)
Solution 2
(a) . Multiply these two binomials and we have reach our answer (remember the formula!)
(b). The original expression is equal to . (Just brute force this out).
(c) Now notice that each term in the sequence is . As seen in part (a), we see that . Follow Solution 1 above.
~hastapasta
See Also
2008 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
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 |