The value of the sum -n-113in-in-1- where i-1 equalsA- iB- i-1C- -iD- 0

∑_n=1^13 ( i^n + i^n+1) = ( i + i^2 + i^3 + ..... + i^13) + ( i^2 + i^3 + .......+ i^14 ) = i(1 i^13)/1 i + i^2 (1 i^13)/1 i = i(1 i/1 i) + i^2(1 i)/(1 i) ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

The value of ...

Question

The value of the sum $13 \sum n = 1 (i^{n} + i^{n + 1})$ , where i = $\sqrt{- 1}$ equals

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i-1

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-i

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Solution

The correct option is B
i-1

$13 \sum n = 1 (i^{n} + i^{n + 1})$ = ( i + $i^{2}$ + $i^{3}$ + ..... + $i^{13}$ ) + ( $i^{2}$ + $i^{3}$ + .......+ $i^{14}$ )

= $\frac{i (1 - i^{13})}{1 - i} + \frac{i^{2} (1 - i^{13})}{1 - i}$

= i $(\frac{1 - i}{1 - i}) + \frac{i^{2} (1 - i)}{(1 - i)} = i + i^{2} = i - 1$ .

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The value of the sum 13∑n=1(in+in+1) , where i = √−1 equals

The correct option is B i-1 13∑n=1(in+in+1) = ( i + i2 + i3 + ..... + i13) + ( i2 + i3 + .......+ i14 ) = i(1−i13)1−i+i2(1−i13)1−i = i(1−i1−i)+i2(1−i)(1−i)=i+i2=i−1 .

The value of the sum $13 \sum n = 1 (i^{n} + i^{n + 1})$ , where i = $\sqrt{- 1}$ equals