Question

How do you find the sum of the finite geometric sequence of $\sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}}$?

Answer 1

Firstly, write the formula for the sum of the finite geometric sequence. We should then compare the variables with that of the formula one’s and then substitute in the formula. Keep evaluating to get the final answer, the sum.

Formula used: The sum of a finite geometric progression is given by the formula, $\dfrac{{a(1 - {r^n})}}{{(1 - r)}}$
Where $a$ is the first term,$n$ is the number of terms, and $r$is the ratio between any two consecutive terms.

Complete step-by-step solution:
The given finite geometric sequence is $\sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}}$
The question is to find the sum of the geometric sequence which can be found out by using the formula, $\sum\limits_{j = 1}^n {a{r^{j - 1}}} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$ Where $a$ is the first term,$n$ is the number of terms, and $r$is the ratio between any two consecutive terms.
Now, we compare the variables in the formula with our given geometric sequence.
$\sum\limits_{j = 1}^n {a{r^{j - 1}}} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$
$\Rightarrow \sum\limits_{j = 1}^n {a{r^{j - 1}} = \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} }$
Here, $a = 32;n = 6;r = \dfrac{1}{4}$
On substituting in, $\dfrac{{a(1 - {r^n})}}{{(1 - r)}}$we get,
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{32\left( {1 - {{\left( {\dfrac{1}{4}} \right)}^6}} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
Now we first evaluate the power $6$ function.
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{32\left( {1 - \left( {\dfrac{1}{{{4^6}}}} \right)} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{32\left( {1 - \left( {\dfrac{1}{{4096}}} \right)} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
Now we evaluate the numerator part.
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{32\left( {\dfrac{{4096 - 1}}{{4096}}} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
Progress to subtraction operation in the numerator and then multiply it with $32\;$
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{32\left( {\dfrac{{4095}}{{4096}}} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
$4096\;$ when divided with $32\;$gives $128\;$.
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{\left( {\dfrac{{4095}}{{128}}} \right)}}{{1 - \left( {\dfrac{1}{4}} \right)}}$
Now, simplify the denominator.
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{\left( {\dfrac{{4095}}{{128}}} \right)}}{{\left( {\dfrac{3}{4}} \right)}}$
Evaluate the RHS and then simplify.
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{4095}}{{128}} \times \dfrac{4}{3}$
$128\;$when divided with $4$ we get $32\;$
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{4095}}{{32}} \times \dfrac{1}{3}$
$4095\;$ when divided with $3$ we get $1365\;$
So, on further simplifying we get,
$\Rightarrow \sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}} = \dfrac{{1365}}{{32}}$

$\therefore$The sum of the finite geometric sequence $\sum\limits_{j = 1}^6 {32{{\left( {\dfrac{1}{4}} \right)}^{j - 1}}}$ is $\dfrac{{1365}}{{32}}$.

Note: Whenever we are asked to give the sum of some finite geometric progression, always write the sequence in this format $\sum\limits_{j = 1}^n {a{r^{j - 1}}}$ to easily put the constants into the formula directly. If this step is skipped , we will have to find the first term separately, write the sequence up to $3$ terms to find the ratio of the geometric progression. So, avoid this long method as it may result in mistakes.