trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work



$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$

can be effectively rewritten as

$$\sum_{k=0}^n 3^{n-k} \cdot 2^k = \sum_{k=0}^n 3^{n} \cdot 3^{-k} \cdot 2^k =\sum_{k=0}^n 3^{n} \cdot \frac{2^k}{3^k} = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k$$

This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by

$$\sum_{k=0}^n ar^k = \frac{a(r^{n+1} - 1)}{r-1}$$


$$\sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = \frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$

This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.

Multiply by $3$ on the top and bottom:

$$\frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = \frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3}$$

Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.

$$\frac{3^{n+1} \cdot ((2/3)^{n+1} - 1)}{2 - 3} = \frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2}$$


$$1 - (2/3)^{n+1} = \frac{3^{n+1}}{3^{n+1}} - \frac{2^{n+1}}{3^{n+1}} = \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$

The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with

$$\frac{3^{n+1} \cdot (1 - (2/3)^{n+1})}{3 - 2} = \frac{3^{n+1} \cdot \frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$


$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = \sum_{k=0}^n 3^n \cdot \left( \frac{2}{3} \right)^k = 3^{n+1} - 2^{n+1}$$

  • $\begingroup$ It looks like your sum has an extra $(2/3)^k$ in it. $\endgroup$ – Michael Burr Dec 13 '18 at 3:32
  • $\begingroup$ Where, exactly? $\endgroup$ – Eevee Trainer Dec 13 '18 at 3:36
  • $\begingroup$ Oh, I see. I'll try to figure out how to resolve it. $\endgroup$ – Eevee Trainer Dec 13 '18 at 3:37
  • $\begingroup$ Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn. $\endgroup$ – Eevee Trainer Dec 13 '18 at 3:42

Sum of first n terms of a geometric series is given by:

$$S_n = \frac{a(r^n-1)}{r-1}.$$

Here $a$ is the first term of the series and $r$ is the common ratio.

In your question, $r = \frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).

  • $\begingroup$ yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work $\endgroup$ – Maxfield Dec 13 '18 at 2:45
  • $\begingroup$ Perhaps it would be helpful to identify $a$ and $r$ in the given problem. $\endgroup$ – Michael Burr Dec 13 '18 at 2:46
  • $\begingroup$ as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though. $\endgroup$ – Maxfield Dec 13 '18 at 2:48
  • $\begingroup$ It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit) $\endgroup$ – Nutan Nepal Dec 13 '18 at 2:54

One gets $T(2n)=3^n\sum_{k=0}^n(\frac23)^k=3^n(\frac{1-(\frac23)^{n+1}}{1-\frac23})=3^{n+1}-2^{n+1}$.


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