Question

Find the value of the sum, $\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$, where, $i=\sqrt{-1}$.

Answer 1

Hint: In this problem we will expand the sum till 13 elements as 13 is a small number. Then we will see a general trend in the higher power of i and use the generalizations. The trend and the generalizations help us to reduce the higher powers of i into lower powers of i and thus make the solution simpler.

Complete step-by-step answer:

Given, Sum = $\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$.
First let us look at the generalizations of the higher powers of i.
We know that, $i=\sqrt{-1}$.
Squaring on both sides would give us, ${{i}^{2}}=-1$.
Now multiplying i on both sides again would give us, ${{i}^{3}}=-i$.
Consider, ${{i}^{2}}=-1$ again.
Squaring on both sides would give us, ${{i}^{4}}=1$.
Thus, we got to know that,
$i=\sqrt{-1}$, ${{i}^{2}}=-1$, ${{i}^{3}}=-i$ and ${{i}^{4}}=1$.
Now, let us observe who to write ${{i}^{5}},{{i}^{6}},{{i}^{7}}$ and ${{i}^{8}}$.
Now consider, ${{i}^{5}}$. ${{i}^{5}}$ can be written as, ${{i}^{4+1}}$.
${{i}^{4+1}}$ can be written as ${{i}^{4}}\times {{i}^{1}}$. But, ${{i}^{4}}=1$ as we saw earlier. Thus, ${{i}^{5}}=i$.
Now, consider ${{i}^{6}}$. ${{i}^{6}}$ can be written as ${{i}^{4+2}}$.
${{i}^{4+2}}$ can be written as ${{i}^{4}}\times {{i}^{2}}$. But, ${{i}^{4}}=1$ and ${{i}^{2}}=-1$.
So, ${{i}^{6}}=\left( 1 \right)\left( -1 \right)=-1$.
Let us now see ${{i}^{7}}$. ${{i}^{7}}$ can be written as ${{i}^{4+3}}$. ${{i}^{4+3}}$ can be written as ${{i}^{4}}\times {{i}^{3}}$. ${{i}^{4}}=1$ and ${{i}^{3}}=-i$ as we saw earlier.
Thus, ${{i}^{7}}=\left( 1 \right)\left( -i \right)=-i$.
Now, ${{i}^{8}}$ can be written as ${{i}^{4+4}}$. ${{i}^{4+4}}$ can be written as ${{i}^{4}}\times {{i}^{4}}$. But, ${{i}^{4}}=1$. Thus, ${{i}^{8}}=\left( 1 \right)\left( 1 \right)=1$.
Therefore to generalize we can write,
${{i}^{4n}}=1$, where, $n\in N$.
Eg: ${{i}^{4}},{{i}^{8}},{{i}^{12}},{{i}^{14}}$, etc.
${{i}^{4n+1}}=i$, where, $n\in N$.
Eg: ${{i}^{1}},{{i}^{5}},{{i}^{9}},{{i}^{13}}$, etc.
${{i}^{4n+2}}=-1$, where, $n\in N$.
Eg: ${{i}^{2}},{{i}^{6}},{{i}^{10}},{{i}^{14}}$, etc.
${{i}^{4n+3}}=-i$, where, $n\in N$.
Eg: ${{i}^{3}},{{i}^{7}},{{i}^{11}},{{i}^{15}}$, etc.
Now knowing the background of generalizations, let us expand the sum.
Sum = $\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$
Sum = $\left( {{i}^{1}}+{{i}^{1+1}} \right)+\left( {{i}^{2}}+{{i}^{2+1}} \right)+\left( {{i}^{3}}+{{i}^{3+1}} \right)+\left( {{i}^{4}}+{{i}^{4+1}} \right)+......+\left( {{i}^{13}}+{{i}^{13+1}} \right)$
Simplifying, we get,
Sum = $\left( i+{{i}^{2}} \right)+\left( {{i}^{2}}+{{i}^{3}} \right)+\left( {{i}^{3}}+{{i}^{4}} \right)+\left( {{i}^{4}}+{{i}^{5}} \right)+......+\left( {{i}^{13}}+{{i}^{14}} \right)$
Now, as we can see, except $i$ and ${{i}^{14}}$, all the elements are repeated. Thus, removing the brackets and grouping, we get,

& =i+\left( {{i}^{2}}+{{i}^{2}} \right)+\left( {{i}^{3}}+{{i}^{3}} \right)+\left( {{i}^{4}}+{{i}^{4}} \right)+\left( {{i}^{5}}+{{i}^{5}} \right)+.....+\left( {{i}^{13}}+{{i}^{13}} \right)+{{i}^{14}} \\\ & =i+2{{i}^{2}}+2{{i}^{3}}+2{{i}^{4}}+2{{i}^{5}}+....+2{{i}^{13}}+{{i}^{14}} \\\ & =i+2\left( {{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)+{{i}^{14}} \\\ \end{aligned}$$ Now, $${{i}^{2}}=-1$$, $${{i}^{3}}=-i$$ and $${{i}^{4}}=1$$as we saw earlier. Now, $${{i}^{5}}$$ can be written as $${{i}^{4\left( 1 \right)+2}}$$, which is of the form $${{i}^{4n+1}}$$, which is -1. Thus, $${{i}^{6}}=-1$$. Now, $${{i}^{7}}$$ can be written as $${{i}^{4\left( 1 \right)+3}}$$, which is of the form $${{i}^{4n+3}}$$, which is $$-i$$. Thus, $${{i}^{7}}=-i$$. Now, $${{i}^{8}}={{i}^{4\left( 2 \right)}}$$, which is of the form $${{i}^{4n}}$$, which is 1. Thus, $${{i}^{8}}=1$$. Similarly, let us list out all the elements. $$\begin{aligned} & {{i}^{9}}={{i}^{4\left( 2 \right)+1}}={{i}^{4n+1}}=i \\\ & {{i}^{10}}={{i}^{4\left( 2 \right)+2}}={{i}^{4n+2}}=-1 \\\ & {{i}^{11}}={{i}^{4\left( 2 \right)+3}}={{i}^{4n+3}}=-i \\\ & {{i}^{12}}={{i}^{4\left( 3 \right)}}={{i}^{4n}}=1 \\\ & {{i}^{13}}={{i}^{4\left( 3 \right)+1}}={{i}^{4n+1}}=i \\\ & {{i}^{14}}={{i}^{4\left( 3 \right)+2}}={{i}^{4n+2}}=-1 \\\ \end{aligned}$$ Thus putting all these values in the sum we get, Sum = $$i+2\left( -1+\left( -i \right)+1+i+\left( -1 \right)+\left( -i \right)+1+i+\left( -1 \right)+\left( -i \right)+1+i \right)+\left( \left( -1 \right) \right)$$ Sum = $$i+2\left( -1-i+1+i-1-i+1+i-1-i+1+i \right)-1$$ Sum = $$i+2\left( 0 \right)-1$$ Sum = $$i-1$$ Thus, the sum = $$\sum\limits_{n=1}^{13}{\left( {{i}^{n}}+{{i}^{n+1}} \right)}$$ = $$i-1$$. Note: You should be very careful with the signs of the complex numbers as in complex numbers squared value comes with a negative which is non – intuitive. This mistake of signs can cause a drastic change in answer. Example: - If you forgot the negative sign in $${{i}^{2}}=1$$ and write $${{i}^{2}}=1$$, then $${{i}^{3}}=i$$ and $${{i}^{4}}=1$$ which is completely wrong. Also apply those generalizations of $${{i}^{4n+1}},{{i}^{4n+2}},{{i}^{4n+3}}$$ and $${{i}^{4n}}$$ only if n is a positive number or zero. If you get a negative power, reciprocate it, write it as a fraction, make the power positive and then apply the generalizations. For example, if the sum was $$\sum\limits_{n=1}^{13}{\left( {{i}^{-n}}+{{i}^{-\left( n+1 \right)}} \right)}$$. We should first make the n, the power positive by writing it as a function, like Sum = $$\sum\limits_{n=1}^{13}{\left( \dfrac{1}{{{i}^{n}}}+\dfrac{1}{{{i}^{n+1}}} \right)}$$ And then apply the generalizations.