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

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 $n$

& \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{aligned}$$ As we have given that $$i=\sqrt{-1}$$ $$\begin{aligned} & \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{aligned}$$ So that we can observe that $$\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{aligned} & \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{aligned}$$ $$\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{aligned} & \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{aligned}$$ So that we can observe that $$\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{aligned} & \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{aligned}$$ Hence, the value of the sum $$\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.