Demonstrate $3+9+27+\cdots +3^{n}=(\frac{3}{2})\left ( 3^{n}-1 \right )$

Question

Good day! I have this for this one, but I cannot get it. What am I doing wrong?

$3+9+27+\cdots +3^{n}=(\cfrac{3}{2})\left ( 3^{n}-1 \right )$

$3^{1}=\left ( \cfrac{3}{2} \right )\left ( 3^{1} -1\right )$

$3=\left ( \cfrac{3}{2} \right )\left ( 2\right )$

$3 = 3$

Then the second part:

$3+9+27+\cdots +3^{k}=(\cfrac{3}{2})\left ( 3^{k}-1 \right )$

$3+9+27+\cdots +3^{k}+3^{k+1}=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )$

In the left side I have:

$3+9+27+\cdots +3^{k}+3^{k+1}=(3+9+27+\cdots +3^{k})+3^{k+1}$

Substituting in:

$3+9+27+\cdots +3^{k}=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )$

$=(\cfrac{3}{2})\left ( 3^{k+1}-1 \right )+3^{k+1}$

Answer 1

You can use the fact that $3^i=\frac{3^{i+1}-3^i}{2}$ whenever $i\in\Bbb N$ $$ \begin{aligned}\sum_{i=1}^n 3^i&=\frac{1}{2}\sum_{i=1}^n 3^{i+1}-3^i\\&=\frac{1}{2}\left(\sum_{i=1}^n 3^{i+1}-\sum_{i=1}^n3^i\right)\\&=\frac{1}{2}\left(\sum_{i=2}^{n+1} 3^{i}-\sum_{i=1}^n3^i\right)\\&=\frac{1}{2}\left(3^{n+1}+\sum_{i=2}^{n} 3^{i}-\left[3+\sum_{i=2}^n3^i\right]\right)\\&=\frac{1}{2}(3^{n+1}-3)\\&=\frac{3}{2}(3^{n}-1) \end{aligned}$$

In general, we have that $a^i=\frac{a^{i+1}-a^i}{a-1}$, and, therefore $$\sum_{i=1}^n a^i=\frac{1}{a-1}\sum_{i=1}^n a^{i+1}-a^i=\frac{1}{a-1}(a^{n+1}-a)=\frac{a(a^{n}-1)}{a-1}$$

Answer 2

Here is what you were looking for.

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

Answer 3

Hint:

You should know from high school that the sum of consecutive terms of a geometric series with common ratio $q$ is given by the formula: $$a_m+a_{m+1}+\dots +a_n=\frac{a_{n+1}-a_m}{q-1}\qquad (n>m).$$

Answer 4

$3+9+27+\cdots +3^{n}=$

$3(1 + 3 + 9 + ... + 3^{n-1}) =$

$\frac 32\cdot [2\times (1 + 3 + 9 + ... + 3^{n-1})]$

So we now have to prove that $3^n-1 = 2(1 + 3 + 9 + ... + 3^{n-1})$

There's a trick and an insight and as a hint I will ask you to think about how on earth the $2$ could be related to $3$.

$2 = (3-1)$

So we have to prove:

$(3-1)(1+3+9 + .... + 3^{n-1}) = 3^n-1$.

Which try it out.....

$(3-1)(1+3+9 + .... + 3^{n-1}) = 3^n-1$
$3(1+3+9 + .... + 3^{n-1}) - 1(1+3+9 + .... + 3^{n-1}) =$
$(3 + 9 + 27 + ...... + 3^n) - (1 + 3+ 9 + ..... + 3^{n-1}) = $
$(3+9+27 + 81 + ..... + 3^{n-2}+3^{n-1} + 3^n) - (1 + 3 + 9 + 27+81 + .... +3^{n-2}+3^{n-1}) =\require{cancel}$
$(\cancel 3+\cancel 9+\cancel{27} + \cancel{81} + ..... + \cancel{3^{n-2}}+\cancel{3^{n-1}} + 3^n) - (1 + \cancel 3+\cancel 9+\cancel{27} + \cancel{81} + ..... + \cancel{3^{n-2}}+\cancel{3^{n-1}}) =$
$3^n -1$.