A) \[{{n}_{1}}={{n}_{2}}+1\]
B) \[{{n}_{1}}={{n}_{2}}-1\]
C) \[{{n}_{1}}={{n}_{2}}\]
D) \[{{n}_{1}}>0,{{n}_{2}}>0\]
Correct Answer: D
Solution :
Using \[{{i}^{3}}=-i,{{i}^{5}}=i\] and\[{{i}^{7}}=-i\], we can write the given expression as \[{{(1+i)}^{{{n}_{1}}}}+{{(1-i)}^{{{n}_{1}}}}+{{(1+i)}^{{{n}_{2}}}}+{{(1-i)}^{{{n}_{2}}}}\] \[=2{{[}^{{{n}_{1}}}}{{C}_{0}}{{+}^{{{n}_{1}}}}{{C}_{2}}{{i}^{2}}{{+}^{{{n}_{1}}}}{{C}_{4}}{{i}^{4}}+.....]\]\[+2{{[}^{{{n}_{2}}}}{{C}_{0}}{{+}^{{{n}_{2}}}}{{C}_{2}}{{i}^{2}}{{+}^{{{n}_{2}}}}{{C}_{4}}{{i}^{4}}+.....]\] \[=2{{[}^{{{n}_{1}}}}{{C}_{0}}{{-}^{{{n}_{1}}}}{{C}_{2}}{{+}^{{{n}_{1}}}}{{C}_{4}}+....]\] \[+2{{[}^{{{n}_{2}}}}{{C}_{0}}{{-}^{{{n}_{2}}}}{{C}_{2}}{{+}^{{{n}_{2}}}}{{C}_{4}}+....]\] This is a real number irrespective of the values of \[{{n}_{1}}\]and \[{{n}_{2}}\].You need to login to perform this action.
You will be redirected in
3 sec