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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