( $i^{10} + 1) (i^{9} + 1) (i^{8} + 1) . . . . . . . (i + 1)$ equal to

-1

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Solution

The correct option is C 0
$(i^{10} + 1) (i^{9} + 1) (i^{8} + 1) . . . . . . . (i + 1)$
$= (i^{10} + 1) (i^{9} + 1) (i^{8} + 1) . . . . . . . (i^{3} + 1) (i^{2} + 1) (i + 1)$
$= (i^{10} + 1) (i^{9} + 1) (i^{8} + 1) . . . . . . . (i^{3} + 1) (- 1 + 1) (i + 1) [∵ i^{2} = - 1]$
$= (i^{10} + 1) (i^{9} + 1) (i^{8} + 1) . . . . . . . (i^{3} + 1) (0) (i + 1) = 0$

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Similar questions

{(\frac{1 - i}{1 + i})}^{10} + {(\frac{1 + i}{1 - i})}^{8} =

i^{2} = - 1

(i^{10} - \frac{1}{i^{11}}) + (i^{11} - \frac{1}{i^{12}}) + (i^{12} - \frac{1}{i^{13}}) + (i^{13} - \frac{1}{i^{14}}) + (i^{14} + \frac{1}{i^{15}})

Express $(1 + i)^{10}$ in the standard form a + ib.

I_{n} = \int t a n^{n} x d x, t h e n I_{0} + I_{1} + 2 (I_{2} + . . . I_{8}) + I_{9} + I_{10}

I_{n} = \infty \int 0 \frac{d x}{{(x + \sqrt{1 + x^{2}})}^{n}},

n > 1

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