What is the value of (1+2+3+\cdots+64)?
Answer and explanation
Correct answer: 2080
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(1+2+\cdots+64=\frac{64\times65}{2}=32\times65=2080\). Hence, option C is correct. Option A equals \(\frac{63\times64}{2}=2016\), the sum up to 63, so it omits 64. Exam tip: cancel the factor 2 with the even number first to calculate quickly.
Frequently asked questions
What is the correct answer to this question?
2080
Why is this the correct answer?
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(1+2+\cdots+64=\frac{64\times65}{2}=32\times65=2080\). Hence, option C is correct. Option A equals \(\frac{63\times64}{2}=2016\), the sum up to 63, so it omits 64. Exam tip: cancel the factor 2 with the even number first to calculate quickly.
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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