prove for P(n)= (n+2)(n-1)/2 using principle of induction

Question

Any idea how to prove this with induction? If you plug in numerical values for n the LHS is not equal to the RHS. What is going on here?

The question is: 1+2+3+4+---------+=[(n+2)(n-1)]/2

Answer 1

If your propositions is

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

does the LHS equal the RHS?

Try n = 5

$2+3+4+5 = 14 \\ \frac{(5+2)(5-1)}{2} = 14$

But that is not a proof

Proof by induction.

base case n = 2

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

Inductive hypothesis.

Suppose,

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

We must show that

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

Based on the inductive hypothesis

$2+3+\cdots + n + n+1\\ \frac {(n+2)(n-1)}{2} + n+1\\ \frac {n^2 +n - 2 + 2n + 2}{2} \\ \frac {(n +3)(n)}{2}$

QED

Answer 2

This is an example where $p(n)\implies p(n+1)$ but $p(n)$ is false because there is no starting point.