Question

Find the value of (I²+1) (i²+2) (i²+3) ... (i²+n)​

Answer 1

Answer:

1+i

2

+i

4

+i

6

+i

8

+i

10

+i

12

+i

14

+i

16

+i

18

+i

20

=1+(−1)+1+(−1)+1+(−1)+1+(−1)+1+(−1)+1

=1 (Ans)

Formula:

i

2

=−1 and i

4

=1

Answer 2

The value of (i²+1) (i²+2) (i²+3) ... (i²+n) is zero (0).

Step-by-step explanation:

Given: (i²+1) (i²+2) (i²+3) ... (i²+n)

To Find: Value of the given expression

Solution:

• Finding the value of (i²+1) (i²+2) (i²+3) ... (i²+n)

We have, ⇒ (i²+1) (i²+2) (i²+3) ... (i²+n)

Since i² = -1, therefore,

⇒ ((-1)+1) ((-1)+2) ((-1)+3) ... ((-1)+n)

⇒ (1-1) (2-1) (3-1) ... (n-1)

⇒ (0) (2-1) (3-1) ... (n-1)

⇒ 0

Hence, the value of (i²+1) (i²+2) (i²+3) ... (i²+n) is zero (0).