If t 2- t -1-0- then t -1- t 2- t 2-1- t 22- t 3-1- t 32- t 27-1- t 272 is equal toA - 54B- 27C- 18D

T^2+t+1=0 ⇒ t= 1+i √(3)2, 1 i √(3)2 = w , w^2 Let t = w nbsp;⇒1t = 1w = w^2 ∴ t + 1t = w + w^2 = 1 If r is not divisile by 3,then t^r+ 1t^r= w^r+ (w^2)^r= ...

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Standard XII

Mathematics

De-Moivre's Theorem

If t 2+ t +1=...

Question

If $t^{2} + t + 1 = 0$ , then ${(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + {(t^{3} + \frac{1}{t^{3}})}^{2} . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}$ is equal to

54

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27

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18

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Solution

The correct option is A $54$
$t^{2} + t + 1 = 0$
$\Rightarrow t = \frac{- 1 + i \sqrt{3}}{2}, \frac{- 1 - i \sqrt{3}}{2} = w, w^{2}$
Let $t = w \Rightarrow \frac{1}{t} = \frac{1}{w} = w^{2}$
$∴ t + \frac{1}{t} = w + w^{2} = - 1$
If $r$ is not divisile by $3$ ,then
$t^{r} + \frac{1}{t^{r}} = w^{r} + (w^{2})^{r} = w^{r} + w^{2 r} = - 1$
If $r$ is divisible by $3$ ,then
$t^{r} + \frac{1}{t^{r}} = 2$
Hence ${(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + {(t^{3} + \frac{1}{t^{3}})}^{2} . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}$
$= (1 + 1 + 1 + . . . .18 times) + (4 + 4 + 4 + . . . .9 times) = 54$

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Similar questions

Q. If

t^{2} + t + 1 = 0

, then

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + {(t^{3} + \frac{1}{t^{3}})}^{2} . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

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Q. If

t^{2} + t + 1 = 0

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{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

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Q. If

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{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . {(t^{27} + \frac{1}{t^{27}})}^{2} = 54

Q. If

t^{2} + t + 1 = 0,

then the value of the expression

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De-Moivre's Theorem

Standard XII Mathematics

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If t2+t+1=0, then (t+1t)2+(t2+1t2)2+(t3+1t3)2....+(t27+1t27)2 is equal to

The correct option is A 54t2+t+1=0 ⇒t=−1+i√32,−1−i√32=w,w2 Let t=w⇒1t=1w=w2 ∴t+1t=w+w2=−1 If r is not divisile by 3,then tr+1tr=wr+(w2)r=wr+w2r=−1 If r is divisible by 3,then tr+1tr=2 Hence (t+1t)2+(t2+1t2)2+(t3+1t3)2....+(t27+1t27)2 =(1+1+1+....18 times)+(4+4+4+....9 times)=54

If $t^{2} + t + 1 = 0$ , then ${(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + {(t^{3} + \frac{1}{t^{3}})}^{2} . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}$ is equal to