What is the sum of the first (63) natural numbers?
Answer and explanation
Correct answer: 2016
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=63\), we get \(\frac{63\times64}{2}=63\times32=2016\). Hence, option B is correct. An answer such as \(1996\) usually results from an error in multiplication or division. Exam tip: for \(1+2+\cdots+n\), directly use \(\frac{n(n+1)}{2}\).
Frequently asked questions
What is the correct answer to this question?
2016
Why is this the correct answer?
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=63\), we get \(\frac{63\times64}{2}=63\times32=2016\). Hence, option B is correct. An answer such as \(1996\) usually results from an error in multiplication or division. Exam tip: for \(1+2+\cdots+n\), directly 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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