Find the least value of n for which1 + 3 + 32 + ⋯ + 3n−1 > 1000

1.	Find the least value of n for which1 + 3 + 32 + ⋯ + 3n−1 > 1000
Answer» We have, 1 + 3 + 3² + ⋯ + 3ⁿ⁻¹ > 1000 ⇒ 3⁰ + 3¹ + 3¹ + ⋯ 3ⁿ⁻¹ > 1000 ⇒ 3⁰\(\big(\frac{3^n − 1}{3 − 1}\big)\) > 1000 ⇒ 3ⁿ − 1 > 2000 ⇒ 3ⁿ > 2001 Least value of n, which satisfies this inequality is n = 7 (∵ 3⁷ = 2187) Hence, least value of n = 7.

Answer»

We have,

1 + 3 + 3² + ⋯ + 3ⁿ⁻¹ > 1000

⇒ 3⁰ + 3¹ + 3¹ + ⋯ 3ⁿ⁻¹ > 1000

⇒ 3⁰\(\big(\frac{3^n − 1}{3 − 1}\big)\) > 1000

⇒ 3ⁿ − 1 > 2000

⇒ 3ⁿ > 2001

Least value of n, which satisfies this inequality is n = 7 (∵ 3⁷ = 2187)

Hence, least value of n = 7.

Discussion