The least value of n, such that (1 + 3 + 3² + .....

Series ⇒ 1 + 3 + 3² +...+ 3ⁿ
It is a geometric series whose common ratio is 3.
As we know that ,

a + ar + ar² + ...... + ar^n–1 =	a(rⁿ - 1)	where , r > 1
	r - 1

Here , a = 1 , r = 3 / 1 = 3² / 3 = 3

∴ 1 + 3 + 3² +...... + 3ⁿ = S_n =	1(3^{n + 1} - 1)
	3 - 1

S_n =	3^{n + 1} - 1
	2

According to question,

∴	3^{n + 1} - 1	> 2000
	2

⇒ 3ⁿ⁺¹ – 1 > 4000
⇒ 3ⁿ⁺¹ > 4000 + 1 = 4001
For n = 7 , 3⁸ = 6561 > 4001