Question

The value of the sum \[\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}\] where \[i=\sqrt{-1}\] equals
1) \[i\]
2) \[i-1\]
3) \[-i\]
4) \[0\]

Answer 1

Hint: In this question we have to use the value of \[i\] which is defined to be equal to \[\sqrt{-1}\]. By using the basic rules of indices we can find the values of different powers of \[i\]. Here, first we simplify the summation for the given values of \[n\] and then by substituting the values of different powers of \[i\] and by simplifying the expression further we can obtain the required result.

Complete step-by-step solution:
Now we have to find the value of the sum \[\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}\] where \[i=\sqrt{-1}\]
For this let us consider the expression and simplify it by substituting the values of \[n\]
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)} \\
 & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+\cdots \cdots +{{i}^{13}} \right)+\left( {{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+\cdots \cdots +{{i}^{14}} \right) \\
\end{align}\]
As we have given that \[i=\sqrt{-1}\]
\[\begin{align}
  & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}={{i}^{2}}\cdot i=-i \\
 & \Rightarrow {{i}^{4}}={{i}^{2}}\cdot {{i}^{2}}=1 \\
 & \Rightarrow {{i}^{5}}={{i}^{4}}\cdot i=i \\
\end{align}\]
So that we can observe that
\[\Rightarrow i=\sqrt{-1},{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1,{{i}^{5}}=i\]
Continuing in this way we can write,
\[\Rightarrow {{i}^{13}}={{i}^{4}}\cdot {{i}^{4}}\cdot {{i}^{4}}\cdot i=i\]
\[\Rightarrow {{i}^{14}}={{i}^{4}}\cdot {{i}^{4}}\cdot {{i}^{4}}\cdot {{i}^{2}}=-1\]
That is the values gets repeated from \[{{i}^{5}}\] onwards and hence the above expression becomes
\[\Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( i+\left( -1 \right)+\left( -i \right)+\left( 1 \right)+\cdots \cdots +\left( i \right) \right)+\left( \left( -1 \right)+\left( -i \right)+1+i+\cdots \cdots +\left( -1 \right) \right)\]
By cancelling the terms having opposite signs we will get
\[\Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=i-1\]
Hence, the value of the sum \[\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}\] where \[i=\sqrt{-1}\] equals to \[\left( i-1 \right)\]
Thus, option (2) is the correct option.

Note:In this type of question students have to remember to find out the values of different powers of \[i\]. Also students must be familiar with the simplification of summation. One of the students may solve the question as follows:
Let us consider the expression
\[\Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}\]
Now we can rewrite \[{{i}^{n+1}}\] as \[{{i}^{n+1}}={{i}^{n}}\cdot i\]
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n}}\cdot i \right)} \\
 & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\sum\limits_{n=1}^{13}{{{i}^{n}}\left( 1+i \right)} \\
\end{align}\]
\[\Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( 1+i \right)\sum\limits_{n=1}^{13}{{{i}^{n}}}\]
Now by substituting the values of \[n\], we can simplify the summation as
\[\Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( 1+i \right)\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+\cdots \cdots +{{i}^{13}} \right)\]
As we have given that \[i=\sqrt{-1}\]
\[\begin{align}
  & \Rightarrow {{i}^{2}}=-1 \\
 & \Rightarrow {{i}^{3}}={{i}^{2}}\cdot i=-i \\
 & \Rightarrow {{i}^{4}}={{i}^{2}}\cdot {{i}^{2}}=1 \\
 & \Rightarrow {{i}^{5}}={{i}^{4}}\cdot i=i \\
\end{align}\]
So that we can observe that
\[\Rightarrow i=\sqrt{-1},{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1,{{i}^{5}}=i\]
Continuing in this way we can write,
\[\Rightarrow {{i}^{13}}={{i}^{4}}\cdot {{i}^{4}}\cdot {{i}^{4}}\cdot i=i\]
Hence the above expression becomes,
\[\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( 1+i \right)\left( i+\left( -1 \right)+\left( -i \right)+\left( 1 \right)+\cdots \cdots +\left( i \right) \right) \\
 & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( 1+i \right)\left( i \right) \\
 & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( i+{{i}^{2}} \right) \\
 & \Rightarrow \sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}=\left( i-1 \right) \\
\end{align}\]
Hence, the value of the sum \[\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}\] where \[i=\sqrt{-1}\] equals to \[\left( i-1 \right)\]
Thus, option (2) is the correct option.