Here is one way.

Prove by induction:

bn = (3/5)[3^n] +(2/5)[(-2)^n]

Okay.

Let us see first if b1 is true.

Since you found b1 = 1, we see if that is true.

When n = 1,

b1 = (3/5)[3^1] +(2/5)[(-2)^1]

b1 = (3/5)[3] +(2/5)[-2]

b1 = 9/5 -4/5

b1 = 5/5 = 1 ----it is true.

Then let us assume when n = k,

bk = (3/5)[3^k] +(2/5)[(-2)^k] is true.

Then if we can show that when n = k+1,

b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)] is true also,

then we can say that

bn = (3/5)[3^n] +(2/5)[(-2)^n] is really true.

It is proven by mathematical induction.

So,

b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)]

b_(k+1) = (3/5)[3^k *3^1] +(2/5)[(-2)^k *(-2)^1]

b_(k+1) = (3/5)[3^k *3] +(2/5)[(-2)^k *(-2)]

b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)

Now we use your findings.

You found

b2 = 7

b_(2+1) = b3 = 13

Plugging those into the Eq.(i), when k=2,

b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)

b_(2+1) = (9/5)[3^2] -(4/5)[(-2)^2]

b_(2+1) = (9/5)[9] -(4/5)[4]

b_(2+1) = 81/5 -16/5

b_(2+1) = 65/5 = 13

Or,

b3 = 13 ---which is true.

Therefore, bn = (3/5)[3^n] +(2/5)[(-2)^n] is proven.