³/₂, 3, 7, 16, 35, 74, ....
Using the method of substitution
When n = 1, 5 . 2^n-2 - 1 = 5 . 2^1-2 - 1
= 5 x 2^-1 - 1
= 5 x ¹/₂ - 1
= ⁵/₂ - 1 = ³/₂
When n = 1, 5 . 2^n-2 - n
= 5 . 2^2-2 - 2
= 5 x 2⁰ - 2
= 5 x 1 – 2 = 3
When n = 3, 5 . 2^n-2 - n
= 5 . 2^3-2 - 3
= 5 x 2¹ - 3
= 5 x 2 – 3
= 10 -3 = 7
When n = 4, 5 . 2^n-2 - n
= 5 . 2^4-2 - 4
= 5 x 2² - 4
= 5 x 4 – 4
= 20 – 4 = 16
When n = 5, 5 . 2^n-2 - n
= 5 . 2^5-2 - 5
= 5 x 2³ - 5
= 5 x 8 – 5
= 40 – 5
= 35
When n = 6, 5 . 2^n-2 - n
= 5 . 2^6-2 - 6
= 5 x 2⁴ - 6
= 5 x 16 – 6
= 80 – 6 = 74
∴ 5 . 2^n-2 - 1 = ³/₂, 3, 7, 16, 35, 74