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Question: If \({{i}^{2}}=-1\) , then sum \(i+{{i}^{2}}+{{i}^{3}}+.........\) to 1000 terms is equal to : a)....

If i2=−1{{i}^{2}}=-1 , then sum i+i2+i3+.........i+{{i}^{2}}+{{i}^{3}}+......... to 1000 terms is equal to :
a). 1
b). -1
c). i
d). 0

Explanation

Solution

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 i2=−1{{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+i2+i3+.........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
Sn=a(1−rn)(1−r){{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,
S1000=i(1−i1000)(1−i){{S}_{1000}}=\dfrac{i\left( 1-{{i}^{1000}} \right)}{\left( 1-i \right)}
Now, since i2=−1{{i}^{2}}=-1 ,
i2×i2=−1×−1 ⇒i4=1 \begin{aligned} & {{i}^{2}}\times {{i}^{2}}=-1\times -1 \\\ & \Rightarrow {{i}^{4}}=1 \\\ \end{aligned}
So, any number of the form i4m{{i}^{4m}} is equal to 1. Since i4m{{i}^{4m}} can be written as (i4)m{{\left( {{i}^{4}} \right)}^{m}} which is (1)m{{\left( 1 \right)}^{m}} which is ultimately 1.
Thus, i1000=(i4)250{{i}^{1000}}={{\left( {{i}^{4}} \right)}^{250}}
i1000=1{{i}^{1000}}=1 …………….. (i)
Therefore,
S1000=i(1−1)(1−i){{S}_{1000}}=\dfrac{i\left( 1-1 \right)}{\left( 1-i \right)} ………… Since i1000=1{{i}^{1000}}=1(from (i))
Which gives us,
S1000=i(0)(1−i){{S}_{1000}}=\dfrac{i\left( 0 \right)}{\left( 1-i \right)}
Which is, S1000=0{{S}_{1000}}=0
Therefore, option (d) is the correct answer.

Note: There is an alternate method to solve this problem. We know that i2=−1{{i}^{2}}=-1 , Thus, i2×i=−i{{i}^{2}}\times i=-i , which is i3=−i{{i}^{3}}=-i .
Now, let us add 1,i,i2,i31,i,{{i}^{2}},{{i}^{3}}
Therefore, 1,i,i2,i31,i,{{i}^{2}},{{i}^{3}}
=1+i+(−1)+(−i) =0 \begin{aligned} & =1+i+\left( -1 \right)+\left( -i \right) \\\ & =0 \\\ \end{aligned}
Therefore, 1+i+i2+i3=01+i+{{i}^{2}}+{{i}^{3}}=0
Now, let us group the series as shown below.
(i+i2+i3+i4)+(i5+i6+i7+i8)+............(i997+i998+i999+i1000)\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, i5{{i}^{5}} from the second and so on up till i997{{i}^{997}} from the last group, as common we get,
i(1+i+i2+i3)+i5(1+i+i2+i3)+............+i997(1+i+i2+i3) =i(0)+i5(0)+............+i997(0) \begin{aligned} & 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{aligned}
=0.