QUESTION 9
Easy
Find the smallest value of \(n\) such that the sum of the first \(n\) natural numbers is greater than 1000.
SOLUTION
1
Recall sum of \(n\) natural numbers formula
\(S_n = \frac{n(n+1)}{2}\)
We need:
\(\frac{n(n+1)}{2} > 1000\)
⇒ \(n(n+1) > 2000\)
⇒ \(n^2 + n − 2000 > 0\)
We need:
\(\frac{n(n+1)}{2} > 1000\)
⇒ \(n(n+1) > 2000\)
⇒ \(n^2 + n − 2000 > 0\)
2
Solve the quadratic equation
\(n^2 + n − 2000 > 0\)
Using the quadratic formula,
\(n = \) \(\begin{aligned} n &= \frac{−1 ± \sqrt{1 + 8000}}{2} \\ &= \frac{−1 ± \sqrt{8001}}{2} \end{aligned}\)
\(\sqrt{8001} ≈ 89.45\)
So \(n ≈ 44.22\)
Since \(n\) must be a whole number and the sum must be greater than 1000,
\(n = 45\)
Using the quadratic formula,
\(n = \) \(\begin{aligned} n &= \frac{−1 ± \sqrt{1 + 8000}}{2} \\ &= \frac{−1 ± \sqrt{8001}}{2} \end{aligned}\)
\(\sqrt{8001} ≈ 89.45\)
So \(n ≈ 44.22\)
Since \(n\) must be a whole number and the sum must be greater than 1000,
\(n = 45\)
🏆
Final Answer : Smallest value of \(n = 45\)
Concept Note
The sum of the first \(n\) natural numbers is
\(S_n = \frac{n(n+1)}{2}\)
To find the smallest \(n\), solve the inequality and then check the nearest whole numbers.