The sum of the first n natural numbers

The sum of the first n natural numbers

Asked by 2 years ago
20 points

The sum of the first n natural numbers is $\frac{n(n+1)}{2}$. Prove this.

The sum

2 Answers

Answered by 2 years ago
393k points

Here is another way of thinking about this. The list of n natural numbers is

$$\mathbb{N} = \{1,2,3,4,5,\ldots,(n-4),(n-3),(n-2),(n-1),n\}$$

now take the sum of the first and last number, and the sum of the second number and second last number, and so on...





$$ \ldots$$


and notice they are all equal to $n+1$. There are $\frac{n}{2}$ such terms and hence the formula


Answered by 2 years ago
365.9k points

Use a proof by induction

Step 1: Show its true for n=1

$$\text{LHS} = 1 \text{, RHS} = \frac{1(1+1)}{2}=1$$

Step 2: Assume its true for n=k


Step 3: Under the assumption of n=k from step 2, show that n=k+1 is true


on the right hand side we should end up with $\frac{(k+1)(k+2)}{2}$ if our assumption from step 2 is correct. Simplify the right hand side





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Views: 18
Asked: 2 years ago