For positive integers ${{n}_{1}},{{n}_{2}}$, the value of the expression ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ is a real number if and only if
(a) ${{n}_{1}}={{n}_{2}}+1$
(b) ${{n}_{1}}={{n}_{2}}-1$
(c) ${{n}_{1}}={{n}_{2}}$
(d) ${{n}_{1}}>0,{{n}_{2}}>0$
Answer
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Hint: Simplify the given expression using the fact that ${{i}^{4n+1}}=i$ and ${{i}^{4n-1}}=-i$ to simplify the terms of the given expression. Use the Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ to expand the terms of the given expression. Simplify the expression by cancelling out the like terms and solve the equation to find the condition for which the expression will be a real number.
Complete step-by-step solution -
We have to find the condition for which the expression ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ is a real number, where ${{n}_{1}},{{n}_{2}}$ are positive integers.
We know that $i=\sqrt{-1}$ is a square root of unity. We also know that ${{i}^{4n+1}}=i$ and ${{i}^{4n-1}}=-i$.
Substituting $n=1$ in the formula ${{i}^{4n-1}}=-i$, we have ${{i}^{4\left( 1 \right)-1}}={{i}^{3}}=-i$.
Substituting $n=1$ in the formula ${{i}^{4n+1}}=i$, we have ${{i}^{4\left( 1 \right)+1}}={{i}^{5}}=i$.
Substituting $n=2$ in the formula ${{i}^{4n-1}}=-i$, we have ${{i}^{4\left( 2 \right)-1}}={{i}^{7}}=-i$.
Thus, we can rewrite the expression ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ as ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}={{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}$.
We will now expand the above expression using the Binomial Theorem.
We know that the Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ .
So, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}$.
Similarly, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}$.
We also have ${{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}$ and ${{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
Thus, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
Simplifying the above expression, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}=2{}^{{{n}_{1}}}{{C}_{0}}-2{}^{{{n}_{1}}}{{C}_{2}}...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+2{}^{{{n}_{2}}}{{C}_{0}}-2{}^{{{n}_{2}}}{{C}_{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
We must have that the above expression is a real number. Thus, we have ${}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}=0$.
So, we have ${}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}\left[ {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}} \right]+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}\left[ {{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}} \right]=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}}=0$ as we know that ${}^{n}{{C}_{n}}=\dfrac{n!}{0!n!}=1$.
Thus, we have ${{\left( i \right)}^{{{n}_{1}}}}+{{\left( -1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{1}}}} \right)+{{\left( i \right)}^{{{n}_{2}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{2}}}} \right)=0$.
So, we must have ${{\left( -1 \right)}^{{{n}_{1}}}}=-1$ and ${{\left( -1 \right)}^{{{n}_{2}}}}=-1$. Thus, ${{n}_{1}}$ and ${{n}_{2}}$ must be both odd numbers. Considering all the options, the given condition holds if and only if ${{n}_{1}}={{n}_{2}}$.
Hence, the condition required for ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ to be a real number is ${{n}_{1}}={{n}_{2}}$, which is option (c).
Note: We can’t solve this question without calculating the higher values of $i=\sqrt{-1}$. We also observe that ${{n}_{1}}$ and ${{n}_{2}}$ can take any positive odd integer as its value. However, based on the given options, we must have ${{n}_{1}}={{n}_{2}}$.
Complete step-by-step solution -
We have to find the condition for which the expression ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ is a real number, where ${{n}_{1}},{{n}_{2}}$ are positive integers.
We know that $i=\sqrt{-1}$ is a square root of unity. We also know that ${{i}^{4n+1}}=i$ and ${{i}^{4n-1}}=-i$.
Substituting $n=1$ in the formula ${{i}^{4n-1}}=-i$, we have ${{i}^{4\left( 1 \right)-1}}={{i}^{3}}=-i$.
Substituting $n=1$ in the formula ${{i}^{4n+1}}=i$, we have ${{i}^{4\left( 1 \right)+1}}={{i}^{5}}=i$.
Substituting $n=2$ in the formula ${{i}^{4n-1}}=-i$, we have ${{i}^{4\left( 2 \right)-1}}={{i}^{7}}=-i$.
Thus, we can rewrite the expression ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ as ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}={{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}$.
We will now expand the above expression using the Binomial Theorem.
We know that the Binomial Theorem which states that for any two numbers ‘x’ and ‘y’, we have ${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$ .
So, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}$.
Similarly, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{1}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{1}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{1}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{1}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{1}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{1}}}}={}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}$.
We also have ${{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}$ and ${{\left( 1+i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}{{\left( 1 \right)}^{{{n}_{2}}}}{{\left( -i \right)}^{0}}+{}^{{{n}_{2}}}{{C}_{1}}{{\left( 1 \right)}^{{{n}_{2}}-1}}{{\left( -i \right)}^{1}}+{}^{{{n}_{2}}}{{C}_{2}}{{\left( 1 \right)}^{{{n}_{2}}-2}}{{\left( -i \right)}^{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( 1 \right)}^{0}}{{\left( -i \right)}^{{{n}_{2}}}}={}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
Thus, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}={}^{{{n}_{1}}}{{C}_{0}}+{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}-{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{0}}-{}^{{{n}_{1}}}{{C}_{1}}i-{}^{{{n}_{1}}}{{C}_{2}}+{}^{{{n}_{1}}}{{C}_{3}}i...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{0}}+{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}-{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{0}}-{}^{{{n}_{2}}}{{C}_{1}}i-{}^{{{n}_{2}}}{{C}_{2}}+{}^{{{n}_{2}}}{{C}_{3}}i...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
Simplifying the above expression, we have ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}=2{}^{{{n}_{1}}}{{C}_{0}}-2{}^{{{n}_{1}}}{{C}_{2}}...+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+2{}^{{{n}_{2}}}{{C}_{0}}-2{}^{{{n}_{2}}}{{C}_{2}}+...+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}$.
We must have that the above expression is a real number. Thus, we have ${}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}{{\left( -i \right)}^{{{n}_{1}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}{{\left( -i \right)}^{{{n}_{2}}}}=0$.
So, we have ${}^{{{n}_{1}}}{{C}_{{{n}_{1}}}}\left[ {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}} \right]+{}^{{{n}_{2}}}{{C}_{{{n}_{2}}}}\left[ {{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}} \right]=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}+{{\left( -i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -i \right)}^{{{n}_{2}}}}=0$ as we know that ${}^{n}{{C}_{n}}=\dfrac{n!}{0!n!}=1$.
Thus, we have ${{\left( i \right)}^{{{n}_{1}}}}+{{\left( -1 \right)}^{{{n}_{1}}}}{{\left( i \right)}^{{{n}_{1}}}}+{{\left( i \right)}^{{{n}_{2}}}}+{{\left( -1 \right)}^{{{n}_{2}}}}{{\left( i \right)}^{{{n}_{2}}}}=0\Rightarrow {{\left( i \right)}^{{{n}_{1}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{1}}}} \right)+{{\left( i \right)}^{{{n}_{2}}}}\left( 1+{{\left( -1 \right)}^{{{n}_{2}}}} \right)=0$.
So, we must have ${{\left( -1 \right)}^{{{n}_{1}}}}=-1$ and ${{\left( -1 \right)}^{{{n}_{2}}}}=-1$. Thus, ${{n}_{1}}$ and ${{n}_{2}}$ must be both odd numbers. Considering all the options, the given condition holds if and only if ${{n}_{1}}={{n}_{2}}$.
Hence, the condition required for ${{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}$ to be a real number is ${{n}_{1}}={{n}_{2}}$, which is option (c).
Note: We can’t solve this question without calculating the higher values of $i=\sqrt{-1}$. We also observe that ${{n}_{1}}$ and ${{n}_{2}}$ can take any positive odd integer as its value. However, based on the given options, we must have ${{n}_{1}}={{n}_{2}}$.
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