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
= $\frac{ar^n–1}{r-1}$, where r >1
Here, a = 1, r = 3
Using the formula: S
n
= $\frac{1(3^{n+1}-1)}{3-1}$
According to the question,
$\frac{3^{n+1}-1}{3-1}$ > 2000
⇒ $3^{n+1}-1$ > 4000
⇒ $3^{n+1}$ > 4001
For $n = 7$
3
8
= 6561 > 4001
Hence, the correct answer is 7.
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