The lowest integer which is greater
than ( 1 + 1/10¹⁰⁰ )^(10¹⁰⁰) is ______________.

Solution

Maths sir · 4 steps

This is (1+1/n)ⁿ with a huge n, so it sits just below e, between 2 and 3.

1. Spot the form. Put n = 10¹⁰⁰. The number is (1+1/n)ⁿ, the same expression whose limit as n → ∞ is e ≈ 2.718. Expand it with the binomial theorem to pin down where it lies.

   (1+1/n)ⁿ = 1 + 1 + 1/2!(1-1/n) + 1/3!(1-1/n)(1-2/n) + …

2. Lower bound. The first two terms give exactly 2, and every term after them is positive. So the number is greater than 2.
3. Upper bound. Each bracket is less than 1, so each term is less than 1/k!. The sum is therefore less than 1 + 1 + 1/2! + 1/3! + … = e, and e < 3.
4. So. 2 < (1+1/10¹⁰⁰)^(10¹⁰⁰) < 3, so the lowest integer greater than it is 3, option (a).

The idea to keep. Whenever you see (1+1/n)ⁿ, remember it always lies strictly between 2 and e for n ≥ 2, and it climbs towards e as n grows.

Answer from the NTA final answer key. Solution by PrepWiser.