What are the first four terms of a sequence given the recursive rule below?

Question

f(n)=2×f(n-1)+3 for all n≥2 and f(1)=7

A. Not enough information

B. 7, 9, 11, 13

C. 7, 15, 31, 63

D. 7, 17, 37, 77

Answer 1

Given equation : f(n)=2×f(n-1)+3 for all n≥2

Given values: f(1)=7

Substitute the given value in the above equation: f(n)=2×f(n-1)+3

f(2)=2* f(2-1)+3

=2* f(1)+3 (Given f(1)=7)

=2* 7 + 3

= 14+3

f(2) = 17

Similarly substitute for n=3 and n=4

f(n)=2×f(n-1)+3

f(3)=2* f(3-1)+3

=2* f(2)+3

=2* 17 + 3

= 34+3

f(3) = 37

n=4:-

f(n)=2×f(n-1)+3

f(4)=2* f(4-1)+3

=2* f(3)+3

=2* 37 + 3

= 74+3

f(4) = 77

So, the values are 7, 17, 37, 77 => option D

Answer 2

In the ilk of SAT, this problem is a "thought" exercise.

Each term is three more that twice the previous.

7, 17, 37, 77.