The value of (1/i)+(1/i2)+(1/i3)+⋅s+(1/i102) is

Question

The value of (1/i)+(1/i2)+(1/i3)+⋅s+(1/i102) is (A) -1 - i (B) 1 + i (C) 1 - i (D) 1 + 2i

Tardigrade Admin · Accepted Answer

(1/i)+(1/i2)+(1/i3)+ ldots+(1/i102) ∴ Sn=((1/i) 1-((1/i))102 /1-((1)i))=((1+1)/i-1) beginpmatrix∵ i4=1 text and i2=-1 endpmatrix =(2(i+1)/i2-1)=-(i+1)=-1-i

Q. The value of $\frac{1}{i} + \frac{1}{i ^{2}} + \frac{1}{i ^{3}} + \dots + \frac{1}{i ^{102}}$ is

$- 1 - i$

$1 + i$

$1 - i$

$1 + 2 i$

$1 - 2 i$

$\frac{1}{i} + \frac{1}{i ^{2}} + \frac{1}{i ^{3}} + \dots + \frac{1}{i ^{102}}$
$∴ S_{n} = \frac{\frac{1}{i} { 1 - ( \frac{1}{i} ) ^{102} }}{1 - ( \frac{1}{i} )} = \frac{( 1 + 1 )}{i - 1}$
$(∵ i^{4} = 1 and i^{2} = - 1)$
$= \frac{2 ( i + 1 )}{i ^{2} - 1} = - (i + 1) = - 1 - i$

Q. The value of i1​+i21​+i31​+⋯+i1021​ is

−1−i

1+i

1−i

1+2i

1−2i

i1​+i21​+i31​+…+i1021​ ∴Sn​=1−(i1​)i1​{1−(i1​)102}​=i−1(1+1)​ (∵i4=1 and i2=−1​) =i2−12(i+1)​=−(i+1)=−1−i