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A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and

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A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and

by Vincen » Wed Jul 22, 2020 12:50 pm

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A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and \(s_2=3.\) Find the value of \(s_6.\)

A. 25
B. 32
C. 36
D. 93
E. 279

[spoiler]OA=E[/spoiler]

Source: Magoosh
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Re: A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and

by Jay@ManhattanReview » Wed Jul 22, 2020 10:56 pm
Vincen wrote: ↑
Wed Jul 22, 2020 12:50 pm
 A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and \(s_2=3.\) Find the value of \(s_6.\)

A. 25
B. 32
C. 36
D. 93
E. 279

[spoiler]OA=E[/spoiler]

Source: Magoosh
Pluggin-in the value of n = 3 in \(s_n=(s_{n-1}-1)(s_{n-2}),\) we get \(s_3=(s_2-1)(s_1)\).

Again, \(s_3=(s_2-1)(s_1) = (3-1)*2=4\)

Similarly, \(s_4=(s_3-1)(s_2) = (4-1)*3=9\)

And, \(s_5=(s_4-1)(s_3) = (9-1)*4=32\)

And finally, \(s_6=(s_5-1)(s_4) = (32-1)*9=279\)

Correct answer: E

Hope this helps!

-Jay
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Re: A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and

by Scott@TargetTestPrep » Sat Jul 25, 2020 2:14 pm
Vincen wrote: ↑
Wed Jul 22, 2020 12:50 pm
 A sequence is defined by \(s_n=(s_{n-1}-1)(s_{n-2})\) for \(n > 2,\) and it has the starting values of \(s_1=2\) and \(s_2=3.\) Find the value of \(s_6.\)

A. 25
B. 32
C. 36
D. 93
E. 279

[spoiler]OA=E[/spoiler]

Solution:

s(3) = (s(2) - 1)(s(1)) = (3 - 1)(2) = 4

s(4) = (s(3) - 1)(s(2)) = (4 - 1)(3) = 9

s(5) = (s(4) - 1)(s(3)) = (9 - 1)(4) = 32

s(6) = (s(5) - 1)(s(4)) = (32 - 1)(9) = 279

Answer: E

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