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The sequence a_n satisfies a_1 = 1 and a_n = 3a_{n - 1} + 1 for n > 1.  Find a_{1978}.

 Dec 15, 2019
 #1
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a_n is just 3 "the times the number before it" + 1.

a_1 = 1

a_2= 1 *3 + 1 = 3+1

a_3  3*(1*3+1)+1   = 3^2 + 3 +1

a_4  3(3*(1*3+1)+1)+1 = 3^3 + 3^2 + 3 + 1

a_5 3(3(3*(1*3+1)+1)+1)+1 = 3^4 + 3^3 + 3^2+ 3+1

...

a_1978 is 3^1977 + 3^1976 + ..... +3 + 1

Hope this helps. 

~~Gf

 Dec 15, 2019
 #2
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+1

The sequence looks like this:
1, 4, 13, 40, 121,.......etc.
The closed form of the sequence is:
a(n) = 1/2 (3^n - 1) 

a(1978) = 1/2 * (3^1978 - 1)

 Dec 15, 2019

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