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

The sum

Susanna

Answered by bhdrkn 2 years ago

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...

$$\overbrace{1,\underbrace{2,\ldots,(n-1)}_{2+(n-1)\,=\,(n+1)},n}^{1+n\,=\,(n+1)}$$

$$a_1+a_n=1+n=(n+1)$$

$$a_2+a_{n-1}=2+(n-1)=(n+1)$$

$$a_3+a_{n-2}=3+(n-2)=(n+1)$$

$$ \ldots$$

$$a_k+a_{n-(k-1)}=k+(n-(k-1))=(n+1)$$

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

$$S_n=\frac{n}{2}(n+1)$$

Answered by JasonJones 2 years ago

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

$$1+2+3+\ldots+k=\frac{k(k+1)}{2}$$

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

$$1+2+3+\ldots+k+(k+1)=\frac{k(k+1)}{2}+(k+1)$$

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

$$\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$$

$$=\frac{k(k+1)+2(k+1)}{2}$$

$$=\frac{(k+1)(k+2)}{2}$$

QED

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Asked: 2 years ago