-1
$\begingroup$

I tried to figure this out by looking at other answers, but mine seems a little more complex and I can't seem to solve it.

$$\sum\limits_{i=0}^{\infty} \frac{3^{i+1}n}{16}4n = 4n\sum\limits_{i=0}^{\infty} \frac{3^{i+1}n}{16} = 4n^2\sum\limits_{i=0}^{\infty} \frac{3^{i+1}}{16}$$

  • 0
    It would become less complex if you index the sum with $k$ instead of $i$. – 2017-01-27
  • 0
    @OpenBall But even then, I still don't understand the steps to solve this. This is what I'm trying to figure out. – 2017-01-27
  • 0
    why don't write $$4n^2\sum_{i=0}^{\infty}\frac{3^{i+1}}{16}$$? – 2017-01-27
  • 0
    @Dr.SonnhardGraubner Good point. But where I do go from there? – 2017-01-27
  • 0
    This sum doesn't converge. See an article about geometric series. – 2017-01-27
  • 1
    Are you trying to write $$\left(\frac3{16}\right)^{i+1}$$ instead of $$\frac{3^{i+1}}{16}\ ?$$ Because then at least the question would make sense (but still be without context). – 2017-01-27

1 Answers 1

0

It simplifies further:

$$3^{i+1}=3\cdot3^i$$

Thus,

$$4n^2\sum\frac{3^{i+1}}{16}=\frac{3n^2}4\sum3^i$$

Now notice that

$$\sum3^i=1+3+9+27+\dots$$

Which does not exist, which means there can not be any closed form.