Question

How can I find the sum of $5-\dfrac{10}{3}+\dfrac{20}{9}-\dfrac{40}{27}+\dfrac{80}{81}-....?$

Answer 1

We assume the sum of the series given in the question to be S. We start to solve the problem by performing arithmetic operations on both sides of the equation until we get the sequence of geometric progression. Then, we use the sum of infinite geometric progression formulas to get the required result.

Complete step by step solution:
In the given question, the number of terms in the series is not countable.
Hence, the given series is infinite.
Let S be the sum of the infinite series given in the question.
$\Rightarrow S=5-\dfrac{10}{3}+\dfrac{20}{9}-\dfrac{40}{27}+\dfrac{80}{81}-....$
Multiplying the above equation with $\left( \dfrac{1}{3} \right)$ on both sides,
$\Rightarrow \dfrac{1}{3}\times S=\dfrac{1}{3}\times \left( 5-\dfrac{10}{3}+\dfrac{20}{9}-\dfrac{40}{27}+\dfrac{80}{81}-.... \right)$
Multiplying the fraction$\left( \dfrac{1}{3} \right)$with each term on the right-hand side, we get,
$\Rightarrow \dfrac{1}{3}S=\dfrac{5}{3}-\dfrac{10}{9}+\dfrac{20}{27}...$
Adding the series $S$ and $\dfrac{1}{3}S$,
$\Rightarrow \dfrac{4}{3}S=5-\dfrac{5}{3}+\dfrac{10}{9}-\dfrac{20}{27}...$
Taking the fraction $-\left( \dfrac{5}{3} \right)$common from the second term on the right-hand side,
$\Rightarrow \dfrac{4}{3}S=5-\dfrac{5}{3}\left( 1-\dfrac{2}{3}+\dfrac{4}{9}... \right)$
$\Rightarrow \dfrac{4}{3}S=5-\dfrac{5}{3}\left( 1-\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}... \right)$
From the above equation,
The expression $\left( 1-\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}... \right)$ is of the form $\left( 1-ar+a{{r}^{2}}... \right)$.
It represents a sum of geometric progression for infinite terms.
The sum of infinite terms in a geometric progression $\left( 1-ar+a{{r}^{2}}... \right)$is given by the formula
$\dfrac{a}{1-r}$
where,
a is the coefficient of each term in the geometric progression.
r is the common ratio between the adjacent terms.
Let G be the sum of geometric progression,
$\Rightarrow G=\left( 1-\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}... \right)$
Here,
$\Rightarrow a=1$
$\Rightarrow r=-\left( \dfrac{2}{3} \right)$
Applying the formula to find the sum of infinite terms of geometric progression,

$\Rightarrow G=\dfrac{1}{\left( 1-\left( -\left( \dfrac{2}{3} \right) \right) \right)}$
Simplifying the above expression, we get,
$\Rightarrow G=\dfrac{1}{\left( 1+\dfrac{2}{3} \right)}$
$\Rightarrow G=\dfrac{1}{\left( \dfrac{5}{3} \right)}$
$\therefore G=\dfrac{3}{5}$
which means,
$\Rightarrow \left( 1-\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}... \right)=\dfrac{3}{5}$
$\Rightarrow \dfrac{4}{3}S=5-\dfrac{5}{3}\left( 1-\dfrac{2}{3}+{{\left( \dfrac{2}{3} \right)}^{2}}... \right)$
Substituting the value of G in the above equation,
$\Rightarrow \dfrac{4}{3}S=5-\dfrac{5}{3}\left( \dfrac{3}{5} \right)$
$\Rightarrow \dfrac{4}{3}S=5-1$
$\Rightarrow \dfrac{4}{3}S=4$
Taking the value of $\left( \dfrac{4}{3} \right)$ to the other side of the equation,
we get,
$\Rightarrow S=\dfrac{\left( 4\times 3 \right)}{4}$
$\therefore S=3$
Substituting the value of S in the above equation,
$\therefore 5-\dfrac{10}{3}+\dfrac{20}{9}-\dfrac{40}{27}+\dfrac{80}{81}-...=3$
Therefore, the sum of the terms for the given infinite series is 3.

Note: It is very important to follow the law of equations while solving any numerical equation. The law states that whatever you do to one side of the equation, the same should be applied to the other side. The given terms are said to be in geometric progression if the ratio between any of the two adjacent terms is the same.