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# In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int

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In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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26 Jan 2015, 06:00
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In the sequence a1, a2, …, an, …, $$a_n=n^2$$ for all n>0. For what positive integer n does $$a_{n-2}+a_{n+1}=a_{n+4}$$?

A. 7
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Posts: 5732
Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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26 Jan 2015, 07:22
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ans E...
two methods...
1) the eqn gives us (n-2)^2+(n+1)^2=(n+4)^2, which gives us on simplificstion
n^2-10n-11=0... so n=11 or -1..
given that n is positive, 11 is the ans..
2) second method.. the given equation basically are three sides of right angle triangle with hypotenuse on right side..
even number can be ruled out as ans.. since even^2+odd^2 can not be even^2..
in odd numbers only 11 satisfies the values for a right angle triangle 9,12 and 15.. as 9^2+12^2=15^2
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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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27 Jan 2015, 10:52
1
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given

an = n^2

therefore

(n-2)^2 + (n+1)^2 = (n+4)^2

simplify

n = -1 or 11

E
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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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27 Jan 2015, 23:27
Hi All,

chetan2u found the "secret" to this question - while the prompt is based in Algebra, the 'shortcut' is actually based in Geometry (specifically a multiple of the 3/4/5 right triangle). You do not need to find the shortcut to answer this question though, if you know your Perfect Squares and how to do basic arithmetic.

The notation in the prompt tells us that each term in the sequence is equal to the "number" of the term squared.

So....
A1 = 1^2 = 1
A2 = 2^2 = 4
A3 = 3^2 = 9
Etc.

We're asked to find the value of N that completes the given equation. Since the answers ARE numbers, we can TEST THE ANSWERS until we find the match.

IF N=7
Does the 5th term + 8th term = 11th term????
25 + 64 is NOT 121

IF N = 8
Does the 6th term + 9th term = 12th term????
36 + 81 is NOT 144

IF N = 9
Does the 7th term + 10th term = 13th term???
49 + 100 is NOT 169

IF N = 10
Does the 8th term + 11th term = 14th term????
64 + 121 is NOT 196

We've eliminated 4 answers, so the remaining answer MUST be correct. Here's the proof though:

IF N = 11
Does the 9th term + 12th term = 15th term???

81 + 144 DOES = 225

[Reveal] Spoiler:
E

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a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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02 Feb 2015, 21:23

$$(n-2)^2 + (n-1)^2 = (n+4)^2$$

$$(n-2+n+4)(n-2-n-4) + (n-1)^2 = 0$$

$$-(2n-2)6 + (n-1)^2 = 0$$

n = 11
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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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02 Feb 2015, 23:26
Bunuel wrote:
In the sequence a1, a2, …, an, …, $$a_n=n^2$$ for all n>0. For what positive integer n does $$a_{n-2}+a_{n+1}=a_{n+4}$$?

A. 7
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.

Just expanding both sides we get (n-2)^2 + (n+1)^2 = (n+4)^2
Expanding and cancelling terms we get n^2-10n-11 = 0
Or (n-11)(n+1) = 0
So n=11 or -1.
Since n is positive, n=11.

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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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27 Apr 2015, 18:58
(n-2)^2 +(n-1)^2=(n-4)^2
(n-4)^2-(n-2)^2=(n-1)^2
(n-4-n+2)(n-4+n-2)=(n-1)^2
-2(2n-6)=(n-1)^2
n=11
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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int [#permalink]

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30 Dec 2017, 02:26
Bunuel wrote:
In the sequence a1, a2, …, an, …, $$a_n=n^2$$ for all n>0. For what positive integer n does $$a_{n-2}+a_{n+1}=a_{n+4}$$?

A. 7
B. 8
C. 9
D. 10
E. 11

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

One method to this problem is starting with n=7 and just plowing through the arithmetic of the answer choices. (Does 5^2 + 8^2 = 11^2? Nope. Does 6^2 + 9^2 = 12^2? Nope. And so forth.) This can get a bit tedious, though.

Algebra offers a better way. Since $$a_{n−2}=(n–2)^2$$, $$a_{n+1}=(n+1)^2$$, and $$a_{n+4}=(n+4)^2$$, the statement $$a_{n−2}+a_{n+1}=a_{n+4}$$ can be rewritten as $$(n–2)^2+(n+1)^2=(n+4)^2$$. Expanding these common algebraic equations gives

$$(n^2–4n+4)+(n^2+2n+1)=n^2+8n+16$$

Simplifying gives

$$2n^2−2n+5=n^2+8n+16$$

Moving everything to the left side produces the quadratic equation

$$n^2–10n–11=0$$

Factoring this equation produces

$$(n–11)(n+1)=0$$

So n = 11 or –1. We know that n is a positive integer, so it must be that n = 11.

Better still, enterprising students may note the 9, 12, 15. Pythagorean triple lurking in this problem. Identifying the a^2 + b^2 = c^2 structure of this problem and searching for a triple that fits can drastically shortcut the work required here.
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Re: In the sequence a1,a2,…,an,…, an=n2 for all n>0. For what positive int   [#permalink] 30 Dec 2017, 02:26
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