Question

Below are the first six terms of a sequence: a1 = 8 a2 = 22 a3 = 50 a4 = 106 a5 = 218 a6 = 442 Fill in the appropriate constant integer values to complete the recursive formula for the sequence.

Answer 1

Check the forward differences of the sequence:

22 - 8 = 14

50 - 22 = 28 = 2*14

106 - 50 = 56 = 4*14

218 - 106 = 112 = 8*14

442 - 218 = 224 = 16*14

The differences are the products of increasing powers of 2 and 14:

$a_2-a_1=14\cdot2^0$

$a_3-a_2=14\cdot2^1$

$a_4-a_3=14\cdot2^2$

and so on, with

$a_n-a_{n-1}=14\cdot2^{n-2}$

$\implies a_n=a_{n-1}+7\cdot2^{n-1}$

Then the sequence has the recursive definition,

$\boxed{\begin{cases}a_1=8\\a_n=a_{n-1}+7\cdot2^{n-1}&\text{for }n>1\end{cases}}$