• # question_answer The value of ${{i}^{1+3+5+...+(2n+1)}}$ is [AMU 1999] 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, - 1 if n is even D) i if n is even, - 1 if n is odd

Let $z={{i}^{[1+3+5+....+(2n+1)]}}$ Clearly series is A.P. with common difference = 2 $\because \,{{T}_{n}}=2n-1$and ${{T}_{n+1}}=2n+1$ So, number of terms in A. P. $=n+1$ Now, ${{S}_{n+1}}=\frac{n+1}{2}[2.1+(n+1-1)2]$ $\Rightarrow {{S}_{n+1}}=\frac{n+1}{2}[2+2n]={{(n+1)}^{2}}$ i.e. ${{i}^{{{(n+1)}^{2}}}}$ Now put $n=1,\,2,\,3,\,4,\,5,\,.....$ $n=1,z={{i}^{4}}=1$, $n=2,\,z={{i}^{6}}=-1$,  $n=3,\,z={{i}^{8}}=1$, $n=4,\,z={{i}^{10}}=-1$, $n=5,\,\,z={{i}^{12}}=1\,,........$