∑k=040i^k=x+iy⇒x^100+x^99y+x^242y^2+x^97y^3=

Correct answer: (D) 1. Given: ∑k=040i^k=x+iy ⇒i^0+i+i^2+i^3+i^4+i^5+i^6+i^7+i^8+...+i^37+i^38+i^39+i^40=x+iy ⇒i^0+0=x+iy ⇒x+iy=1 So, x=1, y=0 Therefore, x^100+x^99y+x^242y^2+x^97y^3=1

Q48 / 160 · AP EAMCET 2022 (04 Jul Shift 2) · Mathematics

$\sum_{k = 0}^{40} i^{k} = x + i y \Rightarrow x^{100} + x^{99} y + x^{242} y^{2} + x^{97} y^{3} =$

A
$0$
B
$- 4$
C
$4$
Correct answer
D
$1$

Solution

Given:
$\sum_{k = 0}^{40} i^{k} = x + i y$
$\Rightarrow i^{0} + [(i + i^{2} + i^{3} + i^{4}) + (i^{5} + i^{6} + i^{7} + i^{8}) + . . . + (i^{37} + i^{38} + i^{39} + i^{40})] = x + i y$
$\Rightarrow i^{0} + 0 = x + i y$
$\Rightarrow x + i y = 1$
So, $x = 1, y = 0$
Therefore,
$x^{100} + x^{99} y + x^{242} y^{2} + x^{97} y^{3} = 1$

Hence, the correct answer is option: 'D'.

∑k=040ik=x+iy⇒x100+x99y+x242y2+x97y3=

Solution

Relevant Questions

$\sum_{k = 0}^{40} i^{k} = x + i y \Rightarrow x^{100} + x^{99} y + x^{242} y^{2} + x^{97} y^{3} =$