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In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege

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Bunuel
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In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege [#permalink] New post   06 Mar 2017, 01:45
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In the sequence a1, a2, … an, \(a_n=\frac{n^2∗(n+1)^2}{4}\) for all positive integers n. In terms of n, what is the value of \(a_{n+1}–a_n\)?

A. n+1
B. 8n^2
C. (n+2)^2
D. 9n^2−8n+7
E. (n+1)^3
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Re: In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege [#permalink] New post   06 Mar 2017, 12:02
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an+1–an = [(n+1) ^2 * (n+2)^2]/4 - [(n) ^2 * (n+1)^2]/4
= [(n+1) ^2]/4] [(n+2)^2 - (n) ^2]
= [(n+1) ^2]/4] [(n) ^2 + 4 + 4n - (n) ^2]
= [(n+1) ^2]/4] [4(n+1)]
= (n+1)^3

Answer is E. (n+1)^3
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Re: In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege [#permalink] New post   09 Mar 2017, 17:21
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Bunuel wrote:
In the sequence a1, a2, … an, \(a_n=\frac{n^2∗(n+1)^2}{4}\) for all positive integers n. In terms of n, what is the value of \(a_{n+1}–a_n\)?

A. n+1
B. 8n^2
C. (n+2)^2
D. 9n^2−8n+7
E. (n+1)^3


Since a(n) = [n^2 * (n+1)^2]/4, a(n+1) = [(n+ 1)^2 * (n+2)^2]/4, and thus:

a(n+1) - a(n)

[(n+ 1)^2 * (n+2)^2]/4 - [n^2 * (n+1)^2]/4

Pull out the common factor (n + 1)^2 from each term:

(n + 1)^2[(n + 2)^2 - n^2]/4

Note that the expression in the square brackets is a difference of squares, which can easily be factored:

(n + 1)^2[(n + 2 - n)(n + 2 + n)]/4

(n + 1)^2[2(2n + 2)]/4

(n + 1)^2[2 * 2(n + 1)]/4

(n + 1)^2[4(n + 1)]/4

(n + 1)^2 * (n + 1)

(n + 1)^3

Answer: E
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In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege [#permalink] New post   Updated on: 06 Mar 2017, 18:33
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Let’s ignore the denominator until the end to make it easier
(n+1)² (n+2)² - [n²(n+1)²]
(n²+2n+1)(n²+4n+4) – [n²(n² + 2n + 1)]
n²(n²+4n+4)+2n(n²+4n+4)+ 1(n²+4n+4) – [n⁴ + 2n³ + n²]
n⁴ + 4n³ + 4n² + 2n³ + 8n² + 8n + n² + 4n + 4 - n⁴ - 2n³ - n²
(4n³ + 12n² + 12n + 4)/4
n³ + 3n² + 3n + 1 = (n+1)³

Therefore, E is the answer. Sorry about the first post. I should have checked over my working out.

Originally posted by matthewsmith_89 on 06 Mar 2017, 11:55.
Last edited by matthewsmith_89 on 06 Mar 2017, 18:33, edited 1 time in total.
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Re: In the sequence a1, a2, an, an=n^2(n+1)^2/4 for all positive intege [#permalink] New post   25 Aug 2022, 00:54
I solved this problem by choosing a postive integer value for n, as it is given in the stem(for all positive integers n, meaning n=>1.

Let n=3 then n+1=4

a3= 3^2 * (3+1)^2 / 4 = 9 * 16 / 4 = 9 * 4 = 36



a(3+1)= a4= 4^2 * (4+1)^2 / 4 = 16 * 5^2 / 4 = 16 * 25 / 4 = 4 * 25 = 100


Thus an=a3=36 and a(n+1)= a(3+1)=a4=100

The final Question: a(n+1) - an = a4-a3= 100 - 36 = 64.

Ask youself the following, for which of the following answer choices if we plug 3 for n give us the 64? the only answer choice that does that, is choice E

(n+1)^3 = (3+1)^3 = 4^3 = 64, thus E
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Re: In the sequence a1, a2, an, an=n^2(n+1)^2/4 for all positive intege [#permalink]
25 Aug 2022, 00:54
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