How to prove $\sum_{k=n}^{2n} k = 3 \sum_{k=1}^n k $ with complete induction?

Question

I was trying to prove this. Have problems to "pull apart" the first sum to put it down to the induction base (n - 1). Base and step are otherwise clear. Could you please help me?

Answer 1

Let's do an induction proof, using the hint of @André Nicolas :

• Initialisation : For $n=1$ you have $\sum_{k=1}^2k=3=3\cdot\sum_{k=1}^1k.$
• Heredity : Suppose the result is true for $n\in\mathbb{N}^*$ and let's show it is true for $n+1.$ You have that $$\sum_{k=n+1}^{2(n+1)}k=\sum_{k=1}^{2n}k+2n+1+2n+2-n=3\cdot\sum_{k=1}^nk+3n+3=3\cdot\sum_{k=1}^{n+1}k$$ and you get your result.

Answer 2

A slight variation:

Suppose $\sum_{k=n-1}^{2(n-1)} k - 3 \sum_{k=1}^{n-1} k = 0 $.

Then

$\begin{array}\\ \sum_{k=n}^{2n} k - 3 \sum_{k=1}^{n} k &=(\sum_{k=n-1}^{2(n-1)} k-(n-1)+2n-1+2n - 3 (\sum_{k=1}^{n-1} k+n)\\ &=\sum_{k=n-1}^{2(n-1)} k-\sum_{k=1}^{n-1} k+ (-(n-1)+2n-1+2n - 3 k)\\ &=0+0\\ &=0\\ \end{array} $

and we are done.