Simplify the expression
${{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}$

Hint: We should try to convert all the exponents of $i$ to the first four powers of $i$ (which are $i,{{i}^{2}},{{i}^{3}},{{i}^{4}}$) and then evaluate the expression.

Complete step by step answer:
The value of i is given by $i=\sqrt{-1}$. Thus, we can evaluate its powers as:
\[\begin{align}
  & i=\sqrt{-1} \\
 & {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1 \\
 & {{i}^{3}}={{i}^{2}}\times i=-1\times i=-i \\
 & {{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}={{\left( -1 \right)}^{2}}=1\text{ }.....(1.1) \\
\end{align}\]
To simplify the problem, we will use Euclid’s division lemma which is stated below:
Euclid’s division lemma: For any two integers a and b such that $b\ne 0$, there exist unique integers c and d such that
$a=cb+d$ and $0 \leq d < b$.
Now, as ${{i}^{4}}=1$, any higher power of $i\text{(}{{i}^{n}}\text{ for n4)}$can be written in terms of its first four powers because, n can be written as
$n=4{{k}_{1}}+{{k}_{2}}$ where k2 is less than 4 by the Euclid’s division lemma and by taking n and 4 in place of a and b in eq(1.2). Thus,
${{i}^{n}}={{i}^{(4{{k}_{1}}+{{k}_{2}})}}={{\left( {{i}^{4}} \right)}^{{{k}_{1}}}}\times {{i}^{{{k}_{2}}}}={{1}^{{{k}_{1}}}}\times {{i}^{{{k}_{2}}}}={{i}^{{{k}_{2}}}}$
Thus, we can write: 5=4×+1+1 and thus${{i}^{5}}={{i}^{(4+1)}}={{i}^{4}}\times i=1\times i=i\text{ }.......\text{(1}\text{.3)}$
Thus, by using equation 1.1 and 1.3, we get
${{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}=-1-i+1+i=0$
Thus the expression evaluates to zero when simplified and the answer is ${{i}^{2}}+{{i}^{3}}+{{i}^{4}}+{{i}^{5}}=0$

Note: Alternatively, we could have directly multiplied $i$ to the required number of times to get the values of ${{i}^{5}}$ I.e. ${{i}^{5}}={{\left( \sqrt{-1} \right)}^{5}}=\sqrt{-1\times -1\times -1\times -1\times -1}=\sqrt{-1}=i$
However, in these types of questions, many times higher powers of I are given and thus, manually multiplying i may take a lot of time and may lead to errors. Thus, converting the powers of I to the first three powers of I by the Euclid’s division lemma is a more convenient method and should be the approach while solving these types of questions.