Find the smallest value of such that the sum of the first natural numbers is greater than 1,000.

Question

Homi · Accepted Answer

We need to find the smallest count of natural numbers whose sum exceeds 1000.

Step 1 — Set up the inequality

Let's recall the formula for the sum of the first $n$ natural numbers. We call this sum $S_n$ .

$S_n = \frac{n(n+1)}{2}$

The problem asks for $S_n$ to be greater than 1000. So, we write this as an inequality.

$\frac{n(n+1)}{2} > 1000$

We can multiply both sides by 2.

$n(n+1) > 2000$

$\boxed{n(n+1) > 2000}$

Step 2 — Find the value of n

We need to find the smallest natural number $n$ that satisfies this. Let's estimate the value of $n$ . We know that $n(n+1)$ is roughly $n^2$ . So, $n^2$ should be approximately 2000. Let's find the square root of 2000. We know that $40^2 = 1600$ and $50^2 = 2500$ . Let's try a number close to $\sqrt{2000}$ . We know $44^2 = 1936$ and $45^2 = 2025$ . So, $n$ should be around 44 or 45.

Let's test $n = 44$ . We calculate the sum of the first 44 natural numbers.

$S_{44} = \frac{44(44+1)}{2}$

$S_{44} = \frac{44 \times 45}{2}$

$S_{44} = 22 \times 45$

$S_{44} = 990$

Since 990 is not greater than 1000, $n=44$ is not our answer. Let's test the next natural number, $n = 45$ . We calculate the sum of the first 45 natural numbers.

$S_{45} = \frac{45(45+1)}{2}$

$S_{45} = \frac{45 \times 46}{2}$

$S_{45} = 45 \times 23$

$S_{45} = 1035$

Since 1035 is greater than 1000, $n=45$ satisfies the condition. This is the smallest such value.

$\boxed{n = 45}$

Answer

The smallest value of $n$ is 45.

Question 9

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