Solveeit Logo

Question

Mathematics Question on Series

Let an=i(n+1)2,{{a}_{n}}={{i}^{{{(n+1)}^{2}}}}, where i=1i=\sqrt{-1} and n=1,2,3.....n=1,2,3..... . Then the value of a1+a3+a5+...+a25{{a}_{1}}+{{a}_{3}}+{{a}_{5}}+...+{{a}_{25}} is

A

13

B

13+i13+i

C

13i13-i

D

12

Answer

13

Explanation

Solution

Given, an=i(n+1)2{{a}_{n}}={{i}^{{{(n+1)}^{2}}}}
\therefore a1=i22=1,a2=i32=i,{{a}_{1}}={{i}^{{{2}^{2}}}}=1,{{a}_{2}}={{i}^{{{3}^{2}}}}=i, a3=i42=1,a4=i52=i,{{a}_{3}}={{i}^{{{4}^{2}}}}=1,{{a}_{4}}={{i}^{{{5}^{2}}}}=i, a5=i62=1,..........{{a}_{5}}={{i}^{{{6}^{2}}}}=1,..........
\therefore For all odd values of n, we get the value of an{{a}_{n}} is 1.
\therefore a1+a3+a5+....+a25{{a}_{1}}+{{a}_{3}}+{{a}_{5}}+....+{{a}_{25}}
=1+1+1+......+113=\underbrace{1+1+1+......+1}_{13}
=13=13