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505-555 (Easy)|   Algebra|   Sequences|      
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hb
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n Updated on: 26 Jul 2013, 12:32
Originally posted by hb on 26 Jul 2013, 12:26.
Last edited by Bunuel on 26 Jul 2013, 12:32, edited 2 times in total.
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In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}-a_{1322}\) ?

(A) 2,245
(B) 2,645
(C) 5,290
(D) 5,545
(E) 5,790

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 90
Question: Page 223
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.
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B
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Bunuel
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 26 Jul 2013, 12:31
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hb
In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}\) - \(a_{1322}\) ?

(A) 2,245
(B) 2,645
(c) 5,290
(D) 5,545
(E) 5,790

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 90
Question: Page 223
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.
Show more

\(a_{1323} -a_{1322}=1323^2-1322^2=(1323-1322)(1323+1322)=2645\)

Answer: B.
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 02 Jan 2015, 04:03
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look for pattern

a_2 - a_1=4-1=3

a_3 - a_2=9-4=5

a_4 - a_3=16-9=7

a_5 - a_4=25-16=9

we see that every pair gives as a result the sum of terms, so solution is just

1323+1322=2645

B
General Discussion
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 10 Oct 2016, 22:08
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an = n^2
From this a1323 = 1323*1323
and a1322 = 1322*1322

a1323 can be rewritten as
1323*(1322+1) = 1323*1322 + 1323

When subtracted with a1322, we get 1323 + 1322(1323-1322) = 1323 + 1322 = 2645
This solution is 2645(Option B)
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 10 Jan 2017, 01:51
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[quote="hb"]In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}-a_{1322}\) ?

(A) 2,245
(B) 2,645
(C) 5,290
(D) 5,545
(E) 5,790


from stem

1323^2 - 1322^2 = ?? this is the difference of the squares of 2 successive integers ( thus (x-y) in (x+y)(x-y) = 1) thus the result is (x+y) ... add them together

gives B
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 21 Jan 2017, 05:07
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hb
In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}-a_{1322}\) ?

(A) 2,245
(B) 2,645
(C) 5,290
(D) 5,545
(E) 5,790

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 90
Question: Page 223
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.
Show more

\(a_n=n^2\)
\(a_{1323}−a_{1322}=1323^2−1322^2=(1323−1322)(1323+1322)=2645\)

Hence option B is correct.
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 16 Mar 2017, 15:19
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Bunuel
hb
In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}\) - \(a_{1322}\) ?

(A) 2,245
(B) 2,645
(c) 5,290
(D) 5,545
(E) 5,790

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 90
Question: Page 223
Edition: Third

My Question: Please provide an explanation on how to arrive at the answer.

\(a_{1323} -a_{1322}=1323^2-1322^2=(1323-1322)(1323+1322)=2645\)

Answer: B.
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Nice way to subtract squares
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 17 Mar 2017, 02:20
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a(1323) - a(1322) = 1323^2 - 1322^2 = (1323 - 1322)(1323+1322) = 1 * 2645 = 2645
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 21 Mar 2017, 06:19
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hb
In the infinite sequence \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{n}\),....\(a_{n}\) = \(n^2\). What is \(a_{1323}-a_{1322}\) ?

(A) 2,245
(B) 2,645
(C) 5,290
(D) 5,545
(E) 5,79
Show more

We are given that a(n) = n^2. So:

a(1323) - a(1322) = 1323^2 - 1322^2 = (1323 - 1322)(1323 + 1322) = 1 x 2,645 = 2,645

Answer: B
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 30 Jan 2023, 08:40
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a(1323) = (1323)^2 = (1322 + 1)^2 --> 1)
a(1322) = (1322)^2 --> 2)
ATQ, Substracting 2) from 1) = 1 + 2644 = 2645 (B)
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In the infinite sequence a_{1}, a_{2}, a_{3}, a_{n},....a_{n 01 Jan 2026, 07:25
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