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
Verified588.9k+ views
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 ${{n}_{1}}={{n}_{2}}$.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success
Why are manures considered better than fertilizers class 11 biology CBSE
Find the coordinates of the midpoint of the line segment class 11 maths CBSE
Distinguish between static friction limiting friction class 11 physics CBSE
The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE
The first National Commission on Labour NCL submitted class 11 social science CBSE
Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
What are Quantum numbers Explain the quantum number class 11 chemistry CBSE
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE