The value of \(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) is A.�

1.	The value of \(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) is A. \(\frac{1}{2}(1+i)\)B. \(\frac{1}{2}(1-i)\)C. 1 D. \(\frac{1}{2}\)
Answer» We know that i⁴ⁿ⁺¹ = i i⁴ⁿ⁺² = i² = -1 i⁴ⁿ⁺³ = i³= -i i⁴ⁿ⁺⁴ = i⁴ = 1 i⁵ + i⁶ + i⁷ + i⁸ + i⁹ = i + (-1) + (-i) + 1 + i = i \(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) = \(\frac{1}{1+i}\) =\(\frac{1}{1+i}\times\frac{(1-i)}{(1-i)}\) = \(\frac{1}{2}(1+i)\)

The value of \(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) is A. \(\frac{1}{2}(1+i)\)B. \(\frac{1}{2}(1-i)\)C. 1 D. \(\frac{1}{2}\)

Answer»

We know that

i⁴ⁿ⁺¹ = i

i⁴ⁿ⁺² = i² = -1

i⁴ⁿ⁺³ = i³= -i

i⁴ⁿ⁺⁴ = i⁴ = 1

i⁵ + i⁶ + i⁷ + i⁸ + i⁹ = i + (-1) + (-i) + 1 + i

= i

\(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) = \(\frac{1}{1+i}\)

=\(\frac{1}{1+i}\times\frac{(1-i)}{(1-i)}\)

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