Question

Find the value of the sum$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}$, where $i=\sqrt{-1}$.
(a) $i$
(b) $i-1$
(c) $-i$
(d) 0

Answer 1

Hint: In this question, you can use the concept of power of Iota. Iota is square root of minus 1 means$\sqrt{-1}$. For example, what is the value of Iota's power 3? Iota $i=\sqrt{-1}$ so ${{i}^{2}}$ is $\sqrt{-1}\times \sqrt{-1}=-1$ and hence${{i}^{3}}={{i}^{2}}\times i=-i$.
A series can be represented in a compact form, called summation or sigma notation. The Greek capital letter $\sum$ is used to represent the sum.

Complete step-by-step answer:
Let us consider the given summation,
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=\sum\limits_{n=1}^{13}{(1+i){{i}^{n}}}$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)\sum\limits_{n=1}^{13}{{{i}^{n}}}=(1+i)\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)$
By using the power of iota concept, we get
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)\left( i-1-i+1+i-1-i+1+i-1-i+1+i \right)$
Cancelling the common terms, we get
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)i$
Open the brackets and multiply by I on the right side, we get
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=i+{{i}^{2}}$
We know that, ${{i}^{2}}=-1$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=i-1$
Hence, the value of the sum $\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}$ is $i-1$.
Therefore, the correct option for the given question is option (b)

Note: Alternatively, the question is solved as follows
Let us consider the summation,
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=\sum\limits_{n=1}^{13}{(1+i){{i}^{n}}}$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)\sum\limits_{n=1}^{13}{{{i}^{n}}}=(1+i)\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)\left( i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}+{{i}^{6}}+{{i}^{7}}+{{i}^{8}}+{{i}^{9}}+{{i}^{10}}+{{i}^{11}}+{{i}^{12}}+{{i}^{13}} \right)$
The second term is a geometric progression on right side and we have to find the summation by using the formula
${{S}_{n}}=a\left( \dfrac{{{r}^{n}}-1}{r-1} \right),r>1$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)i\left( \dfrac{{{i}^{13}}-1}{i-1} \right)$
We have ${{i}^{13}}=i$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)i\left( \dfrac{i-1}{i-1} \right)$
Cancelling the same terms , we get
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=(1+i)i$
Open the brackets and multiply by I on the right side, we get
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=i+{{i}^{2}}$
We know that, ${{i}^{2}}=-1$
$\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}=i-1$
Hence, the value of the sum $\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}$ is $i-1$.
Therefore, the correct option for the given question is option (b)