1. ## Sequences mathematical induction

I have been doing sequences and I have no idea how to do this, can anyone help me step by step on how to solve this problem.

Use mathematical induction to prove that each statement is true for every positive integer n.

3 + 6 + 9 + ... + 3n = 3n(n+1)/2

2. Originally Posted by xterminal01
I have been doing sequences and I have no idea how to do this, can anyone help me step by step on how to solve this problem.

Use mathematical induction to prove that each statement is true for every positive integer n.

3 + 6 + 9 + ... + 3n = 3n(n+1)/2
Hello,

I assume that you know that this proof has to be done in 3 steps:

S1: n = 1: $3 = \frac{3 \cdot 1 \cdot(1+1)}{2}=\frac{3 \cdot 2}{2}=3$ . S1 is true.

S2: We assume that 3 + 6 + 9 + ... + 3n = 3n(n+1)/2 is true.

S3: Show that the equation is true for n+1:

$3 + 6 + 9 + ... + 3n + 3(n+1)= \frac{3n(n+1)}{2}+3(n+1)$

$= \frac{3n(n+1)}{2}+3(n+1)=3(n+1)\left(\frac n2 + 1\right) = 3(n+1)\left(\frac{n+2}{2}\right)$

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

Thus 3 + 6 + 9 + ... + 3n = 3n(n+1)/2 is true for n+1 and therefore it is true for all $n \in \mathbb{N}^*$