The expression 1-in-1-in-2 equals A- - i n -1B- i n -1C- -2 i n -1D- 1

The correct option is C 2i^n+1(1+i)^n(1 i)^n 2 =(1+i)^n×(1+i)^n 2(1 i)^n 2×(1+i)^n 2 =(1+i)^2(n 1)(1+1)^n 2 =(1 1+2i)^n 12^n 2 =2i^n 1 =2i^n 1×i^2 1= 2i^ ...

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Standard XII

Mathematics

Greatest Binomial Coefficients

The expressio...

Question

The expression $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ equals

- i^{n + 1}

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i^{n + 1}

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- 2 i^{n + 1}

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Solution

The correct option is C $- 2 i^{n + 1}$
$\frac{(1 + i)^{n}}{(1 - i)^{n - 2}} = \frac{(1 + i)^{n} \times (1 + i)^{n - 2}}{(1 - i)^{n - 2} \times (1 + i)^{n - 2}} = \frac{(1 + i)^{2 (n - 1)}}{(1 + 1)^{n - 2}} = \frac{(1 - 1 + 2 i)^{n - 1}}{2^{n - 2}} = 2 i^{n - 1} = 2 i^{n - 1} \times \frac{i^{2}}{- 1} = - 2 i^{n + 1}$

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The expression (1+i)n(1−i)n−2 equals

The correct option is C −2in+1(1+i)n(1−i)n−2=(1+i)n×(1+i)n−2(1−i)n−2×(1+i)n−2=(1+i)2(n−1)(1+1)n−2=(1−1+2i)n−12n−2=2in−1=2in−1×i2−1=−2in+1

The expression $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ equals