Question

The value of: $1 + {i^2} + {i^4} + .....{i^{20}}$ is?

Answer 1

The given problem requires us to find the sum of a geometric progression whose first few terms and the last term is given to us. For finding out the sum of a geometric series, we need to know the first term, number of terms and the common ratio of that particular geometric series. We can find out the common ratio of a geometric progression by dividing any two consecutive terms of the series. We can also find out the number of terms in the series using the general term formula as ${a_n} = a{r^{n - 1}}$.

Complete step-by-step solution:
Given geometric series: $1 + {i^2} + {i^4} + .....{i^{20}}$
We have to find the sum of this geometric progression.
Here, first term $= a = 1$.
Now, we can find the common ratio by dividing any two consecutive terms.
So, common ratio $= r = \dfrac{{{i^2}}}{1} = {i^2}$
So, $r = {i^2}$ .
Now, we know that the formula of sum of n terms of a geometric progression is ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$.
So, to calculate the sum of n terms of a GP, we need to know the number of terms, common ratio of GP and the first term.
We can find the common ratio by using the formula for the general term of a GP as ${a_n} = a{r^{n - 1}}$.
So, let us assume the nth term of the given GP to be ${i^{20}}$. So, we have,
${a_n} = a{r^{n - 1}} = {i^{20}}$
Now, substituting the values of a and r, we get,
$\Rightarrow 1{\left( {{i^2}} \right)^{n - 1}} = {i^{20}}$
Simplifying the calculations, we get,
$\Rightarrow {\left( {{i^2}} \right)^{n - 1}} = {i^{20}}$
Using the law of exponents and powers ${\left( {{a^x}} \right)^y} = {a^{xy}}$, we get,
$\Rightarrow {i^{2\left( {n - 1} \right)}} = {i^{20}}$
Now, the bases are the same. So, we can equate the powers as,
$\Rightarrow 2\left( {n - 1} \right) = 20$
Dividing both sides by $2$, we get,
$\Rightarrow \left( {n - 1} \right) = \dfrac{{20}}{2} = 10$
Now, adding one to both sides, we get,
$\Rightarrow n = 11$
So, there are a total of eleven terms in the given GP.
Now, we find the sum of the series.
So, we have, ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
Substituting the values of a, r and n, we get,
$\Rightarrow {S_n} = \dfrac{{1\left( {1 - {{\left( {{i^2}} \right)}^{11}}} \right)}}{{\left( {1 - {i^2}} \right)}}$
Now, we know that ${i^2} = - 1$. So, we get,
$\Rightarrow {S_n} = \dfrac{{1\left( {1 - {{\left( { - 1} \right)}^{11}}} \right)}}{{\left( {1 - \left( { - 1} \right)} \right)}}$
Simplifying the calculations, we get,
$\Rightarrow {S_n} = \dfrac{{1\left( {1 - \left( { - 1} \right)} \right)}}{{\left( {1 + 1} \right)}}$
$\Rightarrow {S_n} = \dfrac{{1\left( {1 + 1} \right)}}{2}$
Doing the calculations, we get the sum as,
$\Rightarrow {S_n} = 1$
So, the sum of geometric progression given to us is ${S_n} = 1$ .

Note: Geometric progression is a series where any two consecutive terms are in the same ratio. The common ratio of a geometric series can be calculated by division of any two consecutive terms of the series. The sum of an infinite geometric progression is ${S_\infty } = \dfrac{a}{{1 - r}}$.