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21 Nov 2016, 02:46
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For the infinite series \(a_1\), \(a_2\), ..., \(a_n\), \(a_n=a_{(n−1)}+n\), and \(a_1=1\). If n>4, is \(a_n\) an odd integer?
(1) \(a_{(n−1)}\) is even
(2) \(a_{(n−4)}\) is odd
(1) \(a_{(n−1)}\) is even
(2) \(a_{(n−4)}\) is odd
21 Nov 2016, 05:33
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Bunuel
For the infinite series \(a_1\), \(a_2\), ..., \(a_n\), \(a_n=a_{(n−1)}+n\), and \(a_1=1\). If n>4, is an an odd integer?
(1) \(a_{(n−1)}\) is even
(2) \(a_{(n−4)}\) is odd
(1) \(a_{(n−1)}\) is even
(2) \(a_{(n−4)}\) is odd
You mean "is \(a_n\) an odd integer?"?
\(a_{1}=1\)
\(a_{2}=a_{1}+2=3\)
\(a_{3}=a_{2}+3=6\)
\(a_{4}=a_{3}+4=10\)
\(a_{5}=a_{4}+5=15\)
\(a_{6}=a_{5}+6=21\)
\(a_{7}=a_{6}+7=28\)
\(a_{7}=a_{6}+7=28\)
Solution 1:
(1) \(a_{n-1}\) is even so \(a_{n-1}=a_{n}-n\) is even, so \(a_n\) could be even or odd. Insufficient
(2) \(a_{n-4}\) is odd so \(a_{n-4}=a_{n-3}-(n-3)=a_{n-2}-(2n-5)=a_{n-1}-(3n-6)=a_n-(4n-6)\) is odd, so \(a_n\) is odd. Sufficient
The answer is B.
Solution 2:
\(a_{1}=1\)
\(a_{2}=a_{1}+2=1+2\)
\(a_{3}=a_{2}+3=1+2+3\)
\(a_{4}=a_{3}+4=1+2+3+4\)
\(a_{k}=1+2+ \dots + k=\frac{k(k+1)}{2}\)
(1) \(a_{n-1}=\frac{n(n-1)}{2}\) is even, so \(n(n-1)\) is divisible by 4. Hence, \(n=4k\) or \(n-1=4k\).
If \(n=4k\)we have \(a_n=\frac{n(n+1)}{2}=\frac{4k(4k+1)}{2}=2k(4k+1)\) is even.
If \(n-1=4k\) , we have \(a_n=\frac{n(n+1)}{2}=\frac{(4k+1)(4k+2)}{2}=(4k+1)(2k+1)\) is odd.
Insufficient.
(2) \(a_{n-4}=\frac{(n-4)(n-3)}{2}\) is odd, so \((n-4)(n-3)\) is divisible by 2 and not by 4. Hence, \(n-4=4k+2\) or \(n-3=4k+2\).
If \(n-4=4k+2 \iff n=4k+6\), we have \(a_n=\frac{n(n+1)}{2}=\frac{(4k+6)(4k+7)}{2}=(2k+3)(4k+7)\) is an odd integer.
If \(n-3=4k+2 \iff n=4k+5\), we have \(a_n=\frac{n(n+1)}{2}=\frac{(4k+5)(4k+6)}{2}=(4k+5)(2k+3)\) is an odd integer.
Sufficient.
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06 Feb 2017, 18:48
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Bunuel
For the infinite series \(a_1\), \(a_2\), ..., \(a_n\), \(a_n=a_{n−1}+n\), and \(a_1=1\). If n>4, is \(a_n\) an odd integer?
(1) \(a_{n−1}\) is even
(2) \(a_{n−4}\) is odd
(1) \(a_{n−1}\) is even
(2) \(a_{n−4}\) is odd
Official solution from Veritas Prep.
As with most sequence problems, your most effective strategy is typically to list out the first few terms of the sequence to see if you can establish a pattern or trend. Here you will find:
First Term: 1 (Odd)
Second Term: 1 + 2 = 3 (Odd)
Third Term: 3 + 3 = 6 (Even)
Fourth Term: 6 + 4 = 10 (Even)
Fifth Term: 10 + 5 = 15 (Odd)
Sixth Term: 15 + 6 = 21 (Odd)
Seventh Term: 21 + 7 = 28 (Even)
Eighth Term: 28 + 8 = 36 (Even)
From that, you should see the pattern that the evenness/oddness of terms in this sequence progresses in pairs: Odd, Odd, Even, Even, Odd, Odd, Even, Even... And you can see the reason why. When you're at an even number and add an odd, you of course get an odd sum. And since you just added an odd, the next integer will be even so you'll be adding an even to an odd to get odd. Having just added an even, the next number will be odd so you'll have odd + odd, which is even, and the next number up to add is even. Then the cycle will repeat.
This means, then, that statement 1 is not sufficient. If you're on an even term for \(a_{n−1}\), the pattern shows that it could be the first of two consecutive evens (in which case an would be the second even), or that it could be the second (in which case an would be odd).
Statement 2, however, is sufficient. Given the list Odd, Odd, Even, Even, Odd, Odd, Even, Even, Odd, Odd, Even, Even, Odd, Odd..., you should see that if you're on an Odd value and move four spaces away, you'll always land on another odd value. Therefore, statement 2 is sufficient and the correct answer is B.
