Series using mathematical induction (2)

a4 = a3 +3 = (63*4) + 3 = 255

which doesn't satisfy the condition as 255<363 when n=3

a5 = a4 +3 = (255*4) + 3 = 1023

and when n=4 it satisfies the condition..

$a_{n+1}=4a_{n}+3$

I will prove with induction that $a_{n}=4^n-1$ . For $n=3 \rightarrow a_3=4^3-1 = 63$ . Suppose that for any integer k the equation is demonstrated, we have to prove $a_{k+1} = 4^{k+1}-1$ .

From hypothesis: $a_{k+1} = 4 a_k+3$

From inductive hypothesis $a_{k+1} = 4 (4^k-1)+3 \rightarrow a_{k+1}= 4^{k+1}-1$ as we want.

So, $a_{n+1} > a_n+300 \rightarrow 4^{n+1} -1 > 4^n-1+300 \rightarrow 4^n(4-1) > 300 \rightarrow 4^n > 100$ Hence $n=4$ .

Given conditions an+1 > an + 300 and an+1= 4an + 3 So 4an + 3 > an + 300 => an > 99 Since, a3 is given as 63, So, a4 will be 4*63 + 3 That definitely seems greater than 99.

We can find the explicit formula for $an$ which is $4^n-1$

$n=4$ is obviously the answer