Show that if n is a power of 3, then $\sum_{i=0}^{\log_3n} 3^i = \frac{3n-1}{2}$

Question

Show that if n is a power of 3, then $\sum_{i=0}^{\log_3n} 3^i = \frac{3n-1}{2}$

$\sum_{i=0}^{\log_3n} 3^i =3^0+3^1+3^2+3^3...+3^{log_3n}$

This is a geometric sequence, so I used the geometric sum formula: $S= \frac{a_1 \cdot (q^n-1)}{q-1}$.

The first element is : $ 3^0 = 1$,

The ratio is : $q=3$,

The number of elemnts (my n) is: $log_3n$. if n is a power of 3 , we get : $\log_3n= 3^x, x\in \mathbb N$. $\sum_{i=0}^{\log_3n} 3^i = \frac{3^0\cdot (3^{\log_3 n}-1)}{3-1}= \frac{3^{\log_3n}-1}{2}=\frac {3^\bbox[yellow]{3^x}-1}{2}$.

I can see that my $n$ needs to be $log_3n +1$, but I can't find a way to do so. I thought that because I am using a geometric sum formula and I'm summing from $0 $ to $log_3n$, instead of summing from : $1$, I have to sum $log_3n +1$, but it doesn't sound like a valid argument. Any help is greatly appreciated.

Answer 1

Make a change: $n=3^m$. Then: $$S=\sum_{i=0}^{m} 3^i=\frac{1\cdot (3^{m+1}-1)}{3-1}=\frac{3\cdot 3^m-1}{2}=\frac{3n-1}{2}.$$

Answer 2

Hint: You can use $3^{\log_3{n}} = n$ and in general $a^{\log_a{b}} = b$. Also, your last formula for series is false. it must be $\frac{a_0(q^{n+1}-1)}{q-1}$ instead of $\frac{a_0(q^{n}-1)}{q-1}$.

Answer 3

$$\sum_{i=0}^{\log_3 n} 3^{i} =\frac{3^{\bbox[yellow]{1+}\log_3 n}-1}{3-1} =\frac{3 \cdot 3^{\log_3 n}-1}{2} =\frac{3n-1}{2}.$$

Answer 4

Since $n$ is a power of $3$ so let $k\in\mathbb{N}$ be such that $n=3^k$. so we have $\log_3^n=\log_3^{3^k}=k$ so we have $$ \sum_{i=0}^{\log_3^n}3^i=\sum_{i=0}^{\log_3^{3^k}}3^i =\sum_{i=0}^{k}3^i=1+3+3^2+\cdots+3^k\times\frac{3-1}{3-1} $$

$$=\frac{3+3^2+\cdots +3^k+3^{k+1}-1-3-3^2-\cdots-3^k}{3-1} $$

$$=\frac{3^{k+1}-1}{2}=\frac{3\cdot3^{k}-1}{2}=\frac{3n-1}{2}$$ As you wish.

Answer 5

$\sum_{i=0}^{k}3^i = 3^0+3^1+....3^k$.

$3\sum_{i=0}^{k}3^i = \ \ \ \ \ \ 3^1 +....3^k +3^{k+1}.$

Subtract:

$(3-1)\sum_{i=0}^{k}3^i =3^{k+1}-1$, or

$\sum_{i=0}^{k}3^i =\dfrac{3^{k+1}-1}{2}.$

Since $n= 3^m \rightarrow m=\log_{3} n$:

$\sum_{i=0}^{\log_3 n}3^i = \sum_{i=0}^{m}3^i=$

$\dfrac{3 3^m-1}{2}=\dfrac{3n-1}{2}.$