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06 Mar 2017, 01:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
73% (02:28) correct
27%
(02:27)
wrong
based on 401
sessions
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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
A. n+1
B. 8n^2
C. (n+2)^2
D. 9n^2−8n+7
E. (n+1)^3
Updated on: 06 Mar 2017, 18:33
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.
Last edited by matthewsmith_89 on 06 Mar 2017, 18:33, edited 1 time in total.
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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.
(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.
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
= [(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
