Prove Summation to Some Number $n$

Question

Let $n\in \mathbb{N}$. Can someone help me prove this by induction:

$$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right) }{ 2 } .$$

Answer 1

when $n=0$ then the formula is true.

suppose that $\sum_{i=0}^{n-1} i=\frac{1}{2}(n(n-1))$.

$\sum_{i=0}^{n} i=n+\sum_{i=0}^{n-1} i=n+\frac{1}{2}(n(n-1))=\frac{1}{2}(n(n+1))$

Answer 2

HINT: Induction works like this: you start by the lower most value and put it in the equation to verify it. Next, you assume the equation to be true for some value $k$. Then try to prove that it works for $k+1$ as well.

Using $\sum_{i=1}^{k}i=\frac{k(k+1)}{2}$, can you prove that $\sum_{i=1}^{k+1} i=\frac{(k+1)(k+2)}{2}$ as well?

Answer 3

If $n=1$ Then

$$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$

Now assume $n=k$. Thus

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

If now $n=k+1$

$$\sum_{i=1}^{k+1}{i}=\left[\sum_{i=1}^k{i}\right]+(k+1)=$$

What does the bracketed term equal? Use algebra to solve.

Answer 4

$n=1$ is clear.

Assume $$\sum _{i=1}^{n-1}i=\frac{(n-1)n}{2}.$$ Then $$\sum _{i=1}^{n}i=\left(\sum _{i=1}^{n-1}i\right)+n=\frac{(n-1)n}{2}+n= \frac{(n-1)n+2n}{2}=\frac{(n+1)n}{2}.$$