1. If $$a_1 = 1$$ and $$a_{n+1} - 3a_n + 2 = 4n$$ for every positive integer n, then $$a_{100}$$ equals
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By: anil on 05 May 2019 02.29 am
Using given condition we find $$a_2$$ = 5 and $$a_3$$ = 21 and so on. We see that the numbers are of form $$3^n-(2*n)$$ So for 100 we have $$3^{100}-200$$
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