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The value of i1+3+5+......+(2n+1) is


Options:
A .   i if n is even, - i if n is odd
B .   1 if n is even, - 1 if n is odd
C .   1 if n is odd, i if n is even
D .   i if n is even, - 1 if n is odd
Answer: Option C
:
C

Let z = i[1+3+5+......+(2n+1)]


Clearly series is A.P with common difference = 2 


∵  Tn = 2n-1 And  Tn+1 = 2n+1


So, number of terms in A.P. = n+1


Now, sn+1=n+12 [2.1+(n+1-1)2] 


Sn+1=n+12[2+2n]=(n+1)2 i.e., i(n+1)2


Now put n = 1,2,3,4,5,........


n = 1, z = i4 = 1, n = 2, z = i5  = -1 , 


n = 3, z = i8 = 1, n = 4, z = i10  = -1 , 


n = 5, z = i12 = 1, .........


 



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