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,
\[\begin{align}
  & =i+\left( {{i}^{2}}+{{i}^{2}} \right)+\left( {{i}^{3}}+{{i}^{3}} \right)+\left( {{i}^{4}}+{{i}^{4}} \right)+\left( {{i}^{5}}+{{i}^{5}} \right)+.....+\left( {{i}^{13}}+{{i}^{13}} \right)+{{i}^{14}} \\
 & =i+2{{i}^{2}}+2{{i}^{3}}+2{{i}^{4}}+2{{i}^{5}}+....+2{{i}^{13}}+{{i}^{14}} \\
 & =i+2\left( {{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)+{{i}^{14}} \\
\end{align}\]
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.
\[\begin{align}
  & {{i}^{9}}={{i}^{4\left( 2 \right)+1}}={{i}^{4n+1}}=i \\
 & {{i}^{10}}={{i}^{4\left( 2 \right)+2}}={{i}^{4n+2}}=-1 \\
 & {{i}^{11}}={{i}^{4\left( 2 \right)+3}}={{i}^{4n+3}}=-i \\
 & {{i}^{12}}={{i}^{4\left( 3 \right)}}={{i}^{4n}}=1 \\
 & {{i}^{13}}={{i}^{4\left( 3 \right)+1}}={{i}^{4n+1}}=i \\
 & {{i}^{14}}={{i}^{4\left( 3 \right)+2}}={{i}^{4n+2}}=-1 \\
\end{align}\]
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.