Question

If ${{i}^{2}}=-1$ , then sum $i+{{i}^{2}}+{{i}^{3}}+.........$ to 1000 terms is equal to :
a). 1
b). -1
c). i
d). 0

Answer 1

Hint: First of all the given series is a finite geometric series. So first we will apply the formula to find the sum of finite geometric series. We will find the sum up to 1000 terms and then we will substitute ${{i}^{2}}=-1$ in the sum obtained. That would give us the final answer.

Complete step-by-step solution -
So, the given series is $i+{{i}^{2}}+{{i}^{3}}+.........$to 1000 terms. The first term of the series is i and the ration of the series is also i.
Now, sum of geometric series is
${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$ , where n is the number of terms, r is the ratio of the series and a is the first term of the series.
Applying the formula we get,
${{S}_{1000}}=\dfrac{i\left( 1-{{i}^{1000}} \right)}{\left( 1-i \right)}$
Now, since ${{i}^{2}}=-1$ ,
$\begin{align}
  & {{i}^{2}}\times {{i}^{2}}=-1\times -1 \\
 & \Rightarrow {{i}^{4}}=1 \\
\end{align}$
So, any number of the form ${{i}^{4m}}$ is equal to 1. Since ${{i}^{4m}}$ can be written as ${{\left( {{i}^{4}} \right)}^{m}}$ which is ${{\left( 1 \right)}^{m}}$ which is ultimately 1.
Thus, ${{i}^{1000}}={{\left( {{i}^{4}} \right)}^{250}}$
${{i}^{1000}}=1$ …………….. (i)
Therefore,
${{S}_{1000}}=\dfrac{i\left( 1-1 \right)}{\left( 1-i \right)}$ ………… Since ${{i}^{1000}}=1$(from (i))
Which gives us,
${{S}_{1000}}=\dfrac{i\left( 0 \right)}{\left( 1-i \right)}$
Which is, ${{S}_{1000}}=0$
Therefore, option (d) is the correct answer.

Note: There is an alternate method to solve this problem. We know that ${{i}^{2}}=-1$ , Thus, ${{i}^{2}}\times i=-i$ , which is ${{i}^{3}}=-i$ .
Now, let us add $1,i,{{i}^{2}},{{i}^{3}}$
Therefore, $1,i,{{i}^{2}},{{i}^{3}}$
$\begin{align}
  & =1+i+\left( -1 \right)+\left( -i \right) \\
 & =0 \\
\end{align}$
Therefore, $1+i+{{i}^{2}}+{{i}^{3}}=0$
Now, let us group the series as shown below.
$\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}} \right)+\left( {{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}} \right)+............\left( {{i}^{997}}+{{i}^{998}}+{{i}^{999}}+{{i}^{1000}} \right)$ .
Now, taking out I from first group, ${{i}^{5}}$ from the second and so on up till ${{i}^{997}}$ from the last group, as common we get,
$\begin{align}
  & i\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)+{{i}^{5}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)+............+{{i}^{997}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right) \\
 & =i\left( 0 \right)+{{i}^{5}}\left( 0 \right)+............+{{i}^{997}}\left( 0 \right) \\
\end{align}$
=0.