Question : The least value of n, such that (1 + 3 + 32+.......+3n) exceeds 2000, is:
Option 1: 5
Option 2: 6
Option 3: 7
Option 4: 8
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Correct Answer: 7
Solution :
Series = 1 + 3 + 3
2
+ ...+ 3
n
It is a geometric series whose common ratio is 3.
We know,
a + ar + ar
2
+.......+ar
n–1
=
Here, a = 1, r = 3
Using the formula: S
n
=
According to the question,
⇒
⇒
For
3
8
= 6561 > 4001
Hence, the correct answer is 7.
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