sigma notation

**wintersoltice** · June 24th 2010, 08:07 AM

Find $\sum_{r=4}^{n}(lg3^{r})$

what i did was:

$=(lg 3) \sum_{r=4}^{n}(r)$
$=(lg 3)(\frac{1}{2})(n^{2}+7n)$

i used $\frac{n}{2}[2a+(n-1)d]$ to expand $\sum_{r=4}^{n}(r)$

but the suggested answer is $(lg 3)(\frac{1}{2})(n-3)(n+4)$ ........
where did i go wrong?

**red_dog** · June 24th 2010, 08:22 AM

The sum has n-3 terms, not n terms.

**wintersoltice** · June 24th 2010, 08:30 AM

Originally Posted by red_dog

The sum has n-3 terms, not n terms.

then...
$\frac{n-3}{2}[2(4)+((n-3)+1)]$
=(1/2)(n-3)(n+6)

....i still don't get the (n+4)....@"@

**red_dog** · June 24th 2010, 08:35 AM

Originally Posted by wintersoltice

then...
$\frac{n-3}{2}[2(4)+((n-3)+1)]$
=(1/2)(n-3)(n+6)

....i still don't get the (n+4)....@"@

$\frac{n-3}{2}[2\cdot 4+(n-3-1)]=\frac{(n-3)(n+4)}{2}$

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