Prove that 2 + 5 + 8...+ (3n -1) = (1/2)n(3n+1) for all n $\displaystyle \in $ N.

Seems to only work for the base case. Can this be proved by induction?

Printable View

- Oct 15th 2009, 08:05 AMp00ndawgInduction proof help
Prove that 2 + 5 + 8...+ (3n -1) = (1/2)n(3n+1) for all n $\displaystyle \in $ N.

Seems to only work for the base case. Can this be proved by induction? - Oct 15th 2009, 09:02 AMMush
Another way to write this is as follows:

Prove that $\displaystyle \displaystyle \sum_{i = 1}^{n} 3i-1 = \frac{1}{2} n (3n+1)

\, \, \forall \, n \in \mathcal{N}$

To prove this by induction, see if it works for the base case of $\displaystyle n = 1$

$\displaystyle LHS = \displaystyle \sum_{i = 1}^{1} 3i-1 = 3(1) -1 = 2 $

$\displaystyle RHS = \frac{1}{2}(1)(3(1)+1) = \frac{4}{2} = 2 $

$\displaystyle LHS = RHS $, thus base case holds.

The next step is to ASSUME the rule holds for k. Take it for granted that:

$\displaystyle \displaystyle \sum_{i = 1}^{k} 3i-1 = \frac{1}{2} k (3k+1) $

Now, using this result, try to prove that the rule also holds for k+1.

Here's a tip:

$\displaystyle \displaystyle \sum_{i = 1}^{k+1} 3i-1 = 2+5+8+11+...+(3(k+1)-1) = \sum_{i=1}^{k} (3i-1) + (3(k+1)-1) $