2005 AIME II Problems/Problem 3

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An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Solution 1

Let's call the first term of the original geometric series $a$ and the common ratio $r$, so $2005 = a + ar + ar^2 + \ldots$. Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$. Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$. We know this series has sum $20050 = \frac{a^2}{1 - r^2}$. Dividing this equation by $\frac{a}{1-r}$, we get $10 = \frac a{1 + r}$. Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$, $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$, so the answer is $399 + 403 = \boxed{802}$.

(We know this last fraction is fully reduced by the Euclidean algorithm -- because $4 = 403 - 399$, $\gcd(403, 399) | 4$. But 403 is odd, so $\gcd(403, 399) = 1$.)

Solution 2

We can write the sum of the original series as $a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005$, where the common ratio is equal to $\dfrac{m}{n}$. We can also write the sum of the second series as $a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050$. Using the formula for the sum of an infinite geometric series $S=\dfrac{a}{1-r}$, where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}$, and the second sequence can be written as $\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}$. Dividing $\text{(2)}$ by $\text{(1)}$, we obtain $\dfrac{a\cdot n}{m+n}=10$, which can also be written as $a\cdot n=10(m+n)$. Substitute this value for $a\cdot n$ back into $\text{(1)}$, we obtain $10\cdot \dfrac{n+m}{n-m}=2005$. Dividing both sides by 10 yields $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$, but this does not output integer solutions. However, we can also write $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ as $\dfrac{n+m}{n-m}=\dfrac{802}{4}$. This gives the system of equations $m+n=802$ and $n-m=4$, which does have integer solutions. Our answer is therefore $m+n=\boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$, respectively, which are co-prime).

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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