Question

The value of 1/i + 1/i^2 + 1/i^3 + 1/i^4 + ... + 1/i^102 is

Answer 1

Answer:

-1 - i

Step-by-step explanation:

1/i + 1/i^2 + 1/i^3 + 1/i^4 + ... + 1/i^102 = 4(1/i -1 -1/i +1) + 1/i -1  = -1 - i

Answer 2

Answer:

The correct answer is  $&t=-i-1$.

Step-by-step explanation:

Geometric Progression (GP) exists as a kind of sequence where each succeeding term is produced by multiplying each initial term by a fixed digit, which exists named a common ratio. This progression exists also comprehended as a geometric sequence of numbers that follow a pattern.

Given:

$\frac{1}{i}+\frac{1}{i^{2}}+\frac{1}{i^{3}}+\ldots+\frac{1}{i^{102}}$

To find the value of

$\frac{1}{i}+\frac{1}{i^{2}}+\frac{1}{i^{3}}+\ldots+\frac{1}{i^{102}}$

Step 1

Let, the value of

$t=\frac{1}{i}+\frac{1}{i^{2}}+\ldots \ldots+\frac{1}{i^{102}}$

Simplify,

$i^{102} t=i^{101}+i^{100}+\ldots .+i^{0}$

Step 2

By using the sum of Geometric Progression,

$a=i^{0}=1$

$&r=i$ and

$&n=102$

By equating, we get

$&i^{102} t=\frac{1\left(i^{102}-1\right)}{i-1}$

$=\frac{i^{2}-1}{i-1}$

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

Step 3

Simplifying the above equation,

$&i^{102} t=i+1$

$\Rightarrow i^{2} t=i+1$

Thus, we get  $&t=-i-1$.

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