Question

Find the value of $1 + {i^2} + {i^4} + {i^6} + \,.............. + {i^{2n}}$ where $i = \sqrt { - 1}$ ,n$\in$N.

Answer 1

The given series is in geometric progression whose common ratio $r = \dfrac{{{i^2}}}{1} = {i^2}$. Firstly we have to find the number of terms in this geometric progression. Then by using the formula of summation of geometric progression we have to find the sum of series and then apply the even or odd condition on ‘n’ and we get the required value.

Complete step-by-step answer:
Here, the given series is $1 + {i^2} + {i^4} + {i^6} + \,.............. + {i^{2n}}$
Common ratio $r = {i^2}$
${n^{th}}$ term of the given series is ${i^{2n}}$
first term $a = 1$
now, suppose number of terms in the given series are $k$
by applying formula for ${n^{th}}$term$= a{r^{k - 1}}$
we get ${i^{2n}} = 1 \times {\left( {{i^2}} \right)^{k - 1}}$
$\Rightarrow {\left( {{i^2}} \right)^n} = {\left( {{i^2}} \right)^{k - 1}}$
Since bases are the same, we can equate the power.
$\Rightarrow n = k - 1$
$\therefore k = n + 1$
Number of terms in the series is $n + 1$
Now, formula for sum(S) of n terms of GP is $a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)$
By putting suitable value from the question, in the above formula we get,
$\Rightarrow$S $= 1\left( {\dfrac{{{{\left( {{i^2}} \right)}^{n + 1}} - 1}}{{\left( {{i^2}} \right) - 1}}} \right)$
We know that ${i^2} = - 1$ and put this in the above equation.
$\Rightarrow$S $= \left( {\dfrac{{{{\left( { - 1} \right)}^{n + 1}} - 1}}{{ - 1 - 1}}} \right)$
$\Rightarrow$S $= \left( {\dfrac{{{{\left( { - 1} \right)}^{n + 1}} - 1}}{{ - 2}}} \right)$
Now, apply the condition on n.
If n is even number then,
$\Rightarrow$S $= \left( {\dfrac{{{{\left( { - 1} \right)}^{odd}} - 1}}{{ - 2}}} \right)$
The odd power of $- 1$ gives the value $- 1$ and even power of $- 1$ gives $1$.
$\therefore$ S $= \left( {\dfrac{{ - 2}}{{ - 2}}} \right) = 1$
If n is odd number then,
$\Rightarrow$S $= \left( {\dfrac{{{{\left( { - 1} \right)}^{even}} - 1}}{{ - 2}}} \right)$
$\therefore$ S $\left( {\dfrac{{1 - 1}}{{ - 2}}} \right) = 0$

Thus, the value of the given series is “ $0$ ” if n is an odd number and “ $1$ ” if n is an even number and n must be a natural number.

Note: Alternative method:
It is known that ${i^{2n}} = - 1$ if n is $1$, $3$, $5$ odd number.
And ${i^{2n = 1}} = 1$ if n is $2$,$4$,$6$ even number.
If n the odd the last term will be $- 1$ and hence the summation is
$\Rightarrow 1 + \left( { - 1} \right) + 1 + \left( { - 1} \right) + 1..........\left( { - 1} \right) = 0$
If n is the even the last term will be $1$ and hence summation will be
$\Rightarrow 1 + \left( { - 1} \right) + 1 + \left( { - 1} \right).......... + 1 = 1$