Proof verification induction

Question

I am barely off and I don't know where my mistake is?

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

This holds true for n = 1

We get $1 = \frac{1(2)}{2} \to 1$

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

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

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

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

$k^2 + 3k + 2 = k^2 + 2k +k +2$

$k^2 + 3k + 2 = k^2 +3k +2$

This is my first time doing induction. This appears to be right? Should I also carry the summation notation all the way through? When am I supposed to drop it? Is there anything that should be stated at the end of proof? Any notation?

Answer 1

There are wrong equations in your proof. For example, $k+(k+1)=\frac {(k+1)(k+2)} 2$ is not valid when $k=2$. Here is the correct argument: for $n=1$: LHS is $1$ and RHS is $\frac {(1)(2)} 2=1$ so the result is true. Suppose the result holds for a particular value of $n$. Consider $\sum\limits_{k=1}^{n+1} k =\sum\limits_{k=1}^{n} k +(n+1)=\frac {n(n+1)} 2 +(n+1)=(n+1) (1+\frac n 2)=(n+1) \frac {n+2} 2=\frac {(n+1)(n+2)} 2$ where we used the induction hypothesis for the second equality. This completes the induction argument.

Answer 2

Here's a more correct way of explaining the inductive step, with a shorter computation, thanks to factorisation instead of systematically expanding:

Suppose that, for some $n\ge 1$, we have $$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$ We have to show that for the next integer, $$\sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2}.$$

Now, from the inductive hypothesis, we have $$\sum_{k=1}^{n+1} k = \sum_{k=1}^n k +(n+1)= \frac{n(n+1)}{2}+(n+1)=(n+1)\biggl(\frac n2+1\biggr)=(n+1)\frac{n+2}2,$$ which proves the inductive step.