Question

If $i^2$ = -1, then the value of $\sum\limits_{n=1}^{200}{{{i}^{n}}}$is:
a). 50
b). -50
c). 0
d). 100

Answer 1

Hint: First of all open the summation sign then you will see the G.P. in iota (i). Then use the summation formula of the G.P. ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ where “a” is the first term and r is the common ratio and n is the number of terms.

Complete step-by-step solution -
First of all we are opening the summation $\sum\limits_{n=1}^{200}{{{i}^{n}}}$ as follows:
$i + i^2 + i^3 + i^4 + ------ + i^{200}$
The above sequence is a G.P. in which first term (a) is “i” and the common ratio is also “i”. Common ratio is calculated by dividing any term by its preceding term. So, let us find the common ratio by taking any term $i^3$ and divide by the previous term $i^2$ we get the common ratio as “i”. The numbers of terms in the sequence are 200.
Now, using the formula of summation of G.P. in the above sequence we get,
${{S}_{n}}=\dfrac{i\left( 1-{{\left( i \right)}^{200}} \right)}{1-i}$
We can write $i^{200}$ as $ ({i}^{2})^{100} $ so we are going to put the value of $i^2 = -1$ which is given in the question in $({i}^{2})^{100}$ we get $(-1)^{100}$. And we know that even the power of a negative number is positive. So, the value of $i^{200}$ is 1.
$ \begin{align}
  & {{S}_{n}}=\dfrac{i\left( 1-1 \right)}{1-i} \\
 & \Rightarrow {{S}_{n}}=0 \\
\end{align} $
Hence, the summation of the sequence is 0.
Hence, the correct answer is option (c).

Note: You can find the number of terms in this sequence as follows:
$i + i^2 + i^3 + i^4 + ------ + i^{200}$
There is a method of finding the number of terms from the G.P. formula. Take the last term of this sequence and use the formula of a general term of the G.P.
$\begin{align}
  & {{T}_{n}}=a{{r}^{n-1}} \\
 & \Rightarrow {{i}^{200}}=i{{\left( i \right)}^{n-1}} \\
 & \Rightarrow {{i}^{200}}={{\left( i \right)}^{n}} \\
\end{align}$
From the above equation, n = 200. So, the number of terms in the given sequence is 200.