Proving the sum of the first $n$ natural numbers by induction [duplicate]

I am currently studying proving by induction but I am faced with a problem.

I need to solve by induction the following question.

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

for all $n > 1$.

Any help on how to solve this would be appreciated.

This is what I have done so far.

Show truth for $N = 1$

Left Hand Side = 1

Right Hand Side = $\frac{1}{2} (1) (1+1) = 1$

Suppose truth for $N = k$

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

Proof that the equation is true for $N = k + 1$

$$1 + 2 + 3 + ... + k + (k + 1)$$

Which is Equal To

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

This is where I'm stuck, I don't know what else to do. The answer should be:

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

Which is equal to:

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

Right?

By the way sorry about the formatting, I'm still new.

Solution 1:

Basic algebra is what's causing the problems: you reached the point

$$\frac{1}{2}K\color{red}{(K+1)}+\color{red}{(K+1)}\;\;\;\:(**)$$

Now just factor out the red terms:

$$(**)\;\;\;=\color{red}{(K+1)}\left(\frac{1}{2}K+1\right)=\color{red}{(K+1)}\left(\frac{K+2}{2}\right)=\frac{1}{2}(K+1)(K+2)$$

Solution 2:

Are you familiar with how induction works? We first prove this theorem for $n=1$. This is true because $$1 = \frac{1\cdot 2}{2}$$

Now assume that this is true for all values less than $n$, we try to show that it is true for $n$. We have $$1+2+\cdots+n-1+n = \frac{n\cdot (n-1)}{2}+n\\ =\frac{n\cdot (n-1)+2n}{2} = \frac{n^2+n}{2}= \frac{n(n+1)}{2}$$

So it's true for $n=1$, and if it's true for all values less than some number it's true for that number, which means it's true for all numbers.