help

 

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

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)