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Question Number 170187 by infinityaction last updated on 18/May/22

Commented by MJS_new last updated on 18/May/22

${it}^{'} s a number with 609 digits starting with$ $550355 . . . and ending with . . . 570632$
Commented by infinityaction last updated on 19/May/22
X+1/x=-1 X²+1/x²=-1 X³+1/x³=2 X²+x+1=0 Multiply both side by x-1 We will get X³-1=0 X³=1 X⁴=x X⁵=x² X⁶=x³=1 So we can say X^(3k+1)=x X^(3k+2)=x² X^3k=1 So X^(3k+1) + 1/x^(3k+1) and x^(3k+2) + 1/x^(3k+2) will get cancel because X^(3k+1) + 1/x^(3k+1) will get even power So -1^(any even term)=1 And X^(3k+2) +1/x^(3k+2) will get odd power because 3k+2= odd term So (-1)^( any odd number)=-1 1+(-1)=0 So now basically we just have to find the sum of (x^3k+1/x^3k)^3k So That will be equal to 2^3k ( for k=1,2,3..... 674) So now will use the formula of sum of gp So sum will be 8(8⁶⁷⁴-1)/7
Commented by infinityaction last updated on 19/May/22

$s i r p l e a s e c h e c k m y s o l u t i o n$
	Answered by Rasheed.Sindhi last updated on 18/May/22
	$A Try x^2 +x+1=0 (x−1)(x^2 +x+1)=0 x^3 −1=0 x=1,ω, ω^2 (=1/ω) 1 is the root of x−1=0 ∴ x=ω,(1/ω) Σ_(n=1) ^(2022) (x^n +(1/x^n ))^n =? Σ_(n=1) ^3 (x^n +(1/x^n ))^n ={(ω+ω)^1 +(ω^2 +ω^2 )^2 +(ω^3 +ω^3 )^3 } ={2ω+(2ω^2 )^2 +(2ω^3 )^3 } ={2ω+4ω^4 +8} ={2ω+4ω+8} ={6ω+8} =2(3ω+4) .... Σ_(n=4) ^6 (x^n +(1/x^n ))^n ={(ω+ω)^4 +(ω^2 +ω^2 )^5 +(ω^3 +ω^3 )^6 } ={16ω^4 +32ω^(10) +64ω^(18) } ={16ω+32ω+64} =16(3ω+4) Σ_(n=7) ^9 (x^n +(1/x^n ))^n ={(ω+ω)^7 +(ω^2 +ω^2 )^8 +(ω^3 +ω^3 )^9 } ={(2ω)^7 +(2ω^2 )^8 +(2)^9 } =2^7 {(ω)^7 +(2ω^(16) )+(2)^2 } =2^7 {3ω+4} Σ_(n=1) ^(2022) (x^n +(1/x^n ))^n =(3ω+4){2+2^4 +2^7 +...+2^(673) )+(ω^(674) +ω^(674) )^(674) =(3ω+4){2+2^4 +2^7 +...+2^(673) )+(2ω^(674) )^(674) =(3ω+4){2+2^4 +2^7 +...+2^(673) )+2^(674) (ω^(674) )^(674) =(3ω+4){2+2^4 +2^7 +...+2^(673) )+2^(674) ω ..... ...$
	$A T r y$ $x^{2} + x + 1 = 0$ $(x - 1) (x^{2} + x + 1) = 0$ $x^{3} - 1 = 0$ $x = 1, ω, ω^{2} (= 1 / ω)$ $1 i s t h e r o o t o f x - 1 = 0$ $∴ x = ω, \frac{1}{ω}$ $\underset{n = 1}{\sum^{2022}} {(x^{n} + \frac{1}{x^{n}})}^{n} = ?$ $\underset{n = 1}{\sum^{3}} {(x^{n} + \frac{1}{x^{n}})}^{n}$ $= {{(ω + ω)}^{1} + {(ω^{2} + ω^{2})}^{2} + {(ω^{3} + ω^{3})}^{3}}$ $= {2 ω + {(2 ω^{2})}^{2} + {(2 ω^{3})}^{3}}$ $= {2 ω + 4 ω^{4} + 8}$ $= {2 ω + 4 ω + 8}$ $= {6 ω + 8}$ $= 2 (3 ω + 4)$ $. . . .$ $\underset{n = 4}{\sum^{6}} {(x^{n} + \frac{1}{x^{n}})}^{n}$ $= {{(ω + ω)}^{4} + {(ω^{2} + ω^{2})}^{5} + {(ω^{3} + ω^{3})}^{6}}$ $= {16 ω^{4} + 32 ω^{10} + 64 ω^{18}}$ $= {16 ω + 32 ω + 64}$ $= 16 (3 ω + 4)$ $\underset{n = 7}{\sum^{9}} {(x^{n} + \frac{1}{x^{n}})}^{n}$ $= {{(ω + ω)}^{7} + {(ω^{2} + ω^{2})}^{8} + {(ω^{3} + ω^{3})}^{9}}$ $= {{(2 ω)}^{7} + {(2 ω^{2})}^{8} + {(2)}^{9}}$ $= 2^{7} {{(ω)}^{7} + (2 ω^{16}) + {(2)}^{2}}$ $= 2^{7} {3 ω + 4}$ $\underset{n = 1}{\sum^{2022}} {(x^{n} + \frac{1}{x^{n}})}^{n}$ $= (3 ω + 4) {2 + 2^{4} + 2^{7} + . . . + 2^{673}) + {(ω^{674} + ω^{674})}^{674}$ $= (3 ω + 4) {2 + 2^{4} + 2^{7} + . . . + 2^{673}) + {(2 ω^{674})}^{674}$ $= (3 ω + 4) {2 + 2^{4} + 2^{7} + . . . + 2^{673}) + 2^{674} {(ω^{674})}^{674}$ $= (3 ω + 4) {2 + 2^{4} + 2^{7} + . . . + 2^{673}) + 2^{674} ω$ $. . . . .$ $. . .$
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