a_n=4a_(n-1)+3 if a1 = 2

Question

**** the _ is to indicate the subscript.

Answer 1

you are told what a₁ is; then if n = 2, a₂ = 4a₁ 3 = 4•2 + 3 = 11.

 

So you know the second value in the series; the formula tells you the NEXT value in the series, using n = 3:

a₃ = 4a₂+ 3 = 4•11 + 3 = 47.  Proceed in this manner, RECURSIVELY, to get one term after another.

Answer 2

OK. Sometimes I work these the nuts-and-bolts way, and you may be tempted to do that too. You find several terms in the sequence and figure out what's in common. But that takes almost forever. So try not to have to do that. LOL

 

So look at the formula instead, and see what it gives you.

 

a(n) = 4a(n-1) + 3 and a(1)=2

 

That number you multiply by a(n-1) is the *common ratio*. Every term in your explicit sequence has that common ratio *raised to the index*, or something closely related to the index.

 

The number you add every time is the *common difference*. Every term in your explicit sequence has that common difference *as a multiplier*.

 

So check it out:

 

a(1) = 2 = 3*4^0 - 1

 

3 as a multiplier. 4 raised to a power (the power is the index - 1). and then subtract 1 to allow you to get that first value in the sequence. Boom!

 

a(1) = 3*4^0 - 1 =    2

a(2) = 3*4^1 - 1 =  11

a(3) = 3*4^2 - 1 =  47

...

a(n) = 3*4^(n-1) - 1