What is the sum of the first (96) natural numbers?
Answer and explanation
Correct answer: 4656
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=96\), \(\frac{96\times97}{2}=48\times97=4656\). Hence, option A is correct. A value such as \(4626\) may result from an error in multiplication or addition. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).
Frequently asked questions
What is the correct answer to this question?
4656
Why is this the correct answer?
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=96\), \(\frac{96\times97}{2}=48\times97=4656\). Hence, option A is correct. A value such as \(4626\) may result from an error in multiplication or addition. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.
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