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if-k-0-200-i-k-p-1-50-i-p-x-iy-then-x-y-is-a-0-1-b-1-1-c-2-3-d-4-8-




Question Number 16373 by gourav~ last updated on 21/Jun/17
if Σ_(k=0) ^(200) i^k +Π_(p=1) ^(50) i^p =x+iy then..(x,y)is...  a. (0,1)  b. (1,−1)  c. (2,3)  d. (4,8)
$${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}… \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\ $$
Commented by gourav~ last updated on 21/Jun/17
sir.. plz solve
$${sir}..\:{plz}\:{solve} \\ $$
Answered by liday last updated on 21/Jun/17
Σ_(k=0) ^(200) i^k =i^0 +i^1 +i^2 +i^3 +i^4 ...+i^(200) =(1+i−1−i)×50+i^(200) =1  Π_(p=1) ^(50) i^p =i×i^2 ×...×i^(50) =i^(1+2+...+50) =i^(((1+50)×50)/2) =i^(1275) =i^(318×4+3) =i^3 =−i  (x,y)=(1,−1)
$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}\mathrm{i}^{\mathrm{k}} =\mathrm{i}^{\mathrm{0}} +\mathrm{i}^{\mathrm{1}} +\mathrm{i}^{\mathrm{2}} +\mathrm{i}^{\mathrm{3}} +\mathrm{i}^{\mathrm{4}} …+\mathrm{i}^{\mathrm{200}} =\left(\mathrm{1}+\mathrm{i}−\mathrm{1}−\mathrm{i}\right)×\mathrm{50}+\mathrm{i}^{\mathrm{200}} =\mathrm{1} \\ $$$$\underset{\mathrm{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}\mathrm{i}^{\mathrm{p}} =\mathrm{i}×\mathrm{i}^{\mathrm{2}} ×…×\mathrm{i}^{\mathrm{50}} =\mathrm{i}^{\mathrm{1}+\mathrm{2}+…+\mathrm{50}} =\mathrm{i}^{\frac{\left(\mathrm{1}+\mathrm{50}\right)×\mathrm{50}}{\mathrm{2}}} =\mathrm{i}^{\mathrm{1275}} =\mathrm{i}^{\mathrm{318}×\mathrm{4}+\mathrm{3}} =\mathrm{i}^{\mathrm{3}} =−\mathrm{i} \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{1},−\mathrm{1}\right) \\ $$
Commented by gourav~ last updated on 21/Jun/17
ThNk u..
$${ThNk}\:{u}.. \\ $$

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