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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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Re: In the sequence a1, a2, … an, an=n^2∗(n+1)^2/4 for all positive intege [#permalink]
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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

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Re: In the sequence a1, a2, an, an=n^2(n+1)^2/4 for all positive intege [#permalink]
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
Re: In the sequence a1, a2, an, an=n^2(n+1)^2/4 for all positive intege [#permalink]
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