JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • 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

    Correct Answer: C

    Solution :

    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 \[n=1,\,2,\,3,\,4,\,5,\,.....\] \[n=1,z={{i}^{4}}=1\], \[n=2,\,z={{i}^{6}}=-1\],  \[n=3,\,z={{i}^{8}}=1\], \[n=4,\,z={{i}^{10}}=-1\], \[n=5,\,\,z={{i}^{12}}=1\,,........\]

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