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?
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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.
Balloon
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That you very much - it's much easier to use $n+1$ instead of $n-1$ version... – isindia Nov 29 '15 at 22:03
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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.
marty cohen
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Tried: Base: $n=1$ : $$\sum_{k=1}^{2} k = 1 + 2 = 3 * 1 = 3 \sum_{k=1}^1 k$$ Step: from $n-1$ to $n$: we know that $$\sum_{k=n-1}^{2(n-1)} k = 3 \sum_{k=1}^{n-1} k$$
$$\sum_{k=n}^{2n} k = \sum_{k=n-1}^{2(n-1)} k + 2n-1 + 3n + n = ?????$$
Otherwise I thought that I need something like $$3 \sum_{k=1}^n k = 3 (\sum_{k=1}^(n-1) k + n) = 3 sum_{k=1}^(n-1) k + 3n$$
– isindia Nov 29 '15 at 21:48