A) 13
B) \[13+i\]
C) \[13-i\]
D) 12
E) \[12-i\]
Correct Answer: A
Solution :
Given, \[{{a}_{n}}={{i}^{{{(n+1)}^{2}}}}\] \[\therefore \] \[{{a}_{1}}={{i}^{{{2}^{2}}}}=1,{{a}_{2}}={{i}^{{{3}^{2}}}}=i,\] \[{{a}_{3}}={{i}^{{{4}^{2}}}}=1,{{a}_{4}}={{i}^{{{5}^{2}}}}=i,\] \[{{a}_{5}}={{i}^{{{6}^{2}}}}=1,..........\] \[\therefore \]For all odd values of n, we get the value of \[{{a}_{n}}\]is 1. \[\therefore \] \[{{a}_{1}}+{{a}_{3}}+{{a}_{5}}+....+{{a}_{25}}\] \[=\underbrace{1+1+1+......+1}_{13}\] \[=13\]
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