1. ## sum

hi

Am i correct to think

$\Sigma k^3 = \Sigma k^2 \times k = \Sigma k^2 \times \Sigma k, \forall k$?

2. Originally Posted by 1cookie
Am i correct to think
$\Sigma k^3 = \Sigma k^2 \times k = \Sigma k^2 \times \Sigma k, \forall k$?
$\sum\limits_{k = 1}^3 {k^3 } = 36\;,\;\sum\limits_{k = 1}^3 {k^2 } = 14\;\& \;\sum\limits_{k = 1}^3 k = 6$

Now what do you think?

3. Originally Posted by Plato
$\sum\limits_{k = 1}^3 {k^3 } = 36\;,\;\sum\limits_{k = 1}^3 {k^2 } = 14\;\& \;\sum\limits_{k = 1}^3 k = 6$

Now what do you think?

...hmm, clearly $14 \times 6 \neq 36$

i see your point Plato!