The nth term of the sequence \(\frac{3}{2}\), 3, 7, 16, 35, 74 ….. is
-
5 . 2n-2 - n
- 5 . 2n-2 - \(\frac{(n+1)}{2}\)
- 3 . 2n-2
- \(\frac{3n}{2}\)
The correct answer is: A
Explanation
3/2, 3, 7, 16, 35, 74, ....
Using the method of substitution
When n = 1, 5 . 2n-2 - 1 = 5 . 21-2 - 1
= 5 x 2-1 - 1
= 5 x 1/2 - 1
= 5/2 - 1 = 3/2
When n = 1, 5 . 2n-2 - n
= 5 . 22-2 - 2
= 5 x 20 - 2
= 5 x 1 – 2 = 3
When n = 3, 5 . 2n-2 - n
= 5 . 23-2 - 3
= 5 x 21 - 3
= 5 x 2 – 3
= 10 -3 = 7
When n = 4, 5 . 2n-2 - n
= 5 . 24-2 - 4
= 5 x 22 - 4
= 5 x 4 – 4
= 20 – 4 = 16
When n = 5, 5 . 2n-2 - n
= 5 . 25-2 - 5
= 5 x 23 - 5
= 5 x 8 – 5
= 40 – 5
= 35
When n = 6, 5 . 2n-2 - n
= 5 . 26-2 - 6
= 5 x 24 - 6
= 5 x 16 – 6
= 80 – 6 = 74
∴ 5 . 2n-2 - 1 = 3/2, 3, 7, 16, 35, 74
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