Question

Q.22. Evaluate (i+i2+13+14+15+...+ i2021)
TO

Answer 1

Correct question :-

$\rm i + {i}^{2} + {i}^{3} + {i}^{4}... + {i}^{2021}$

Solution :-

We have to find sum of the given series :-

$\rm i + {i}^{2} + {i}^{3} + {i}^{4}... + {i}^{2021}$

[ where i = √(-1) ]

The above expression is in Geometric progression ( G.P) with common ratio i.

Now, we know that :-

Sum of n terms of a geometric progression series with common ratio r and first term a :-

$\boxed{\rm S = \frac{a(1 - {r}^{n}) }{1 - r} }$

First term of the given G.P = i

Number of terms(n) = 2021

common ratio = i

$\rm \longrightarrow \: S = \dfrac{i(1 - {i}^{2021}) }{1 - i}$

Now first we will find out the value of i²⁰²¹

$\rm \longrightarrow {i}^{2021} = {i}^{2020} \times i \\ \\ \rm \longrightarrow {i}^{2021} = ({i}^{4} )^{505} \times i \\ \\ \rm {i}^{4} = 1\\ \\ \rm \longrightarrow {i}^{2021} = (1 )^{505} \times i\\ \\ \rm \longrightarrow {i}^{2021} = i$

Therefore ,the value of i²⁰²¹ = i

$\rm \longrightarrow \: S = \dfrac{i(1 - {i}) }{(1 - i)}$

${\rm \longrightarrow \: S = i}$

Answer :-

$\rm i + {i}^{2} + {i}^{3} + {i}^{4}... + {i}^{2021} = i$