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For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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21 Nov 2016, 01:46
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Re: For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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21 Nov 2016, 04:33
Bunuel wrote: 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 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) \(a_{n1}\) is even so \(a_{n1}=a_{n}n\) is even, so \(a_n\) could be even or odd. Insufficient (2) \(a_{n4}\) is odd so \(a_{n4}=a_{n3}(n3)=a_{n2}(2n5)=a_{n1}(3n6)=a_n(4n6)\) 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) \(a_{n1}=\frac{n(n1)}{2}\) is even, so \(n(n1)\) is divisible by 4. Hence, \(n=4k\) or \(n1=4k\). If \(n=4k\)we have \(a_n=\frac{n(n+1)}{2}=\frac{4k(4k+1)}{2}=2k(4k+1)\) is even. If \(n1=4k\) , we have \(a_n=\frac{n(n+1)}{2}=\frac{(4k+1)(4k+2)}{2}=(4k+1)(2k+1)\) is odd. Insufficient. (2) \(a_{n4}=\frac{(n4)(n3)}{2}\) is odd, so \((n4)(n3)\) is divisible by 2 and not by 4. Hence, \(n4=4k+2\) or \(n3=4k+2\). If \(n4=4k+2 \iff n=4k+6\), we have \(a_n=\frac{n(n+1)}{2}=\frac{(4k+6)(4k+7)}{2}=(2k+3)(4k+7)\) is an odd integer. If \(n3=4k+2 \iff n=4k+5\), 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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For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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06 Feb 2017, 17:48
Bunuel wrote: 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 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.
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For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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06 Feb 2017, 19:07
Statement 1 : \(a_{(n1)}\) is even.
Given : \(a_n\)  \(a_{(n1)}\) = n
\(a_n\) = n + \(a_{(n1)}\)
\(a_n\) = Odd + Even > Odd if n is odd \(a_n\) = Even + Even > Even if n is Even Not Sufficient .
Statement 2 : \(a_{(n−4)}\) is odd
Given : \(a_n\)  \(a_{(n1)}\) = n
So now we need to relate \(a_{(n−4)}\) with \(a_{(n)}\)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2) \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3) \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)
(1) + (2) +(3)+(4)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2)  \(a_{(n−2)}\)  \(a_{(n−4)}\) = 2n \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3)  \(a_{(n−1)}\)  \(a_{(n−4)}\) = 3n \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)  \(a_{(n)}\) \(a_{(n−4)}\) = 4n 
Now ,
\(a_{(n)}\) = 4n + \(a_{(n−4)}\) \(a_{(n)}\) = Even + Odd = Odd > Sufficient
Answer B.



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Re: For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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07 Feb 2017, 15:39
pranjal123 wrote: Statement 1 : \(a_{(n1)}\) is even.
Given : \(a_n\)  \(a_{(n1)}\) = n
\(a_n\) = n + \(a_{(n1)}\)
\(a_n\) = Odd + Even > Odd if n is odd \(a_n\) = Even + Even > Even if n is Even Not Sufficient .
Statement 2 : \(a_{(n−4)}\) is odd
Given : \(a_n\)  \(a_{(n1)}\) = n
So now we need to relate \(a_{(n−4)}\) with \(a_{(n)}\)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2) \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3) \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)
(1) + (2) +(3)+(4)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2)  \(a_{(n−2)}\)  \(a_{(n−4)}\) = 2n \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3)  \(a_{(n−1)}\)  \(a_{(n−4)}\) = 3n \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)  \(a_{(n)}\) \(a_{(n−4)}\) = 4n 
Now ,
\(a_{(n)}\) = 4n + \(a_{(n−4)}\) \(a_{(n)}\) = Even + Odd = Odd > Sufficient
Answer B. Anyone have a suggestion on what I should study to understand this question? The question doesn't make sense to me, and the answers don't help. I clearly need to build some foundation on this topic. Thanks!



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For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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07 Feb 2017, 16:03
funnypanks wrote: pranjal123 wrote: Statement 1 : \(a_{(n1)}\) is even.
Given : \(a_n\)  \(a_{(n1)}\) = n
\(a_n\) = n + \(a_{(n1)}\)
\(a_n\) = Odd + Even > Odd if n is odd \(a_n\) = Even + Even > Even if n is Even Not Sufficient .
Statement 2 : \(a_{(n−4)}\) is odd
Given : \(a_n\)  \(a_{(n1)}\) = n
So now we need to relate \(a_{(n−4)}\) with \(a_{(n)}\)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2) \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3) \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)
(1) + (2) +(3)+(4)
\(a_{(n−3)}\)  \(a_{(n−4)}\) = n (1) \(a_{(n−2)}\)  \(a_{(n−3)}\) = n (2)  \(a_{(n−2)}\)  \(a_{(n−4)}\) = 2n \(a_{(n−1)}\)  \(a_{(n−2)}\) = n (3)  \(a_{(n−1)}\)  \(a_{(n−4)}\) = 3n \(a_{(n)}\)  \(a_{(n−1)}\) = n (4)  \(a_{(n)}\) \(a_{(n−4)}\) = 4n 
Now ,
\(a_{(n)}\) = 4n + \(a_{(n−4)}\) \(a_{(n)}\) = Even + Odd = Odd > Sufficient
Answer B. Anyone have a suggestion on what I should study to understand this question? The question doesn't make sense to me, and the answers don't help. I clearly need to build some foundation on this topic. Thanks! Well,As the question tags says "Number Properties" so you need to study Number properties topic. As you can see question test 2 properties here : 1: outcome of adding ODD and EVEN numbers. 2: Difference between 2 consecutive numbers is constant. In general, for any series based question you need to find the pattern for small set of numbers. As in this question , Plug n=7 you will get \(a_{7}\)  \(a_{6}\) = n On plugging n=6 \(a_6\)  \(a_{5}\) = n And here you have it . 2 numbers have a common difference. With this you can see the pattern in above 2 equation and move forward . OR Just to make things more clearer Put n =11 everywhere in the question and then solve it. Hope this helps



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Re: For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4
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28 Jul 2018, 07:34
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Re: For the infinite series a1, a2, ..., an, an=a(n−1)+n, and a1=1. If n>4 &nbs
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