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List the first 9 terms of the sequence defined recursively by s_(n)=s_(n-2)*(s_(n-1)-1), with s_(1)=2 an s_(2)=3

User BStateham
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Answer:

The first 9 terms of the sequence are:

2, 3, 4, 9, 32, 279, 8928, 2490912, 22237084672.

Explanation:

: s_n = s_(n-2) * (s_(n-1) - 1), where s_1 = 2 and s_2 = 3. This means that to find each term in the sequence, you can use the values of the two previous terms.

s_1 is given as 2.

s_2 is also given as 3.

To find s_3, you use s_1 and s_2: s_3 = s_1 * (s_2 - 1) = 2 * (3 - 1) = 4.

To find s_4, you use s_2 and s_3: s_4 = s_2 * (s_3 - 1) = 3 * (4 - 1) = 9.

To find s_5, you use s_3 and s_4: s_5 = s_3 * (s_4 - 1) = 4 * (9 - 1) = 32.

similarly for next sequence the answer is

s6 = 279

s7 = 8928

s8 = 2490912

s9 = 22237084672

Therefore The first 9 terms of the sequence are:

2, 3, 4, 9, 32, 279, 8928, 2490912, 22237084672.

User RobbeM
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