# If a_n = i ^(n+1)2, where i =√-1 and n = 1,2,3... Then , the value of a1 + a3 +a5+...+a25 is (1) 13 (2) 13 + i (3) 13 - i (4) 12 (5) 12 - i

Solution:

Given an = i (n+1)2

a1 + a3 +a5+…+a25 =$$i^{(1+1)^{2}}+i^{(3+1)^{2}}+i^{(5+1)^{2}}+..i^{(25+1)^{2}}$$

There are 13 odd numbers from 1 to 25.

=i4 +i16+i36+…i676

= 1+1+1+…1 (since i4 = 1)

= 13

Hence option (1) is the answer.

5 (1)

(1)
(1)

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