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Question Number 87933 by john santu last updated on 07/Apr/20
how many terms are there  in (x^2 +x+1)^(11)  ?
howmanytermsaretherein(x2+x+1)11?
Commented by mr W last updated on 07/Apr/20
x^0 ... x^(22)   ⇒23 terms
x0x2223terms
Commented by mr W last updated on 07/Apr/20
but how many terms are there in  (x^5 +x^2 +1)^(11)  ?
buthowmanytermsaretherein(x5+x2+1)11?
Commented by john santu last updated on 07/Apr/20
56 terms
56terms
Commented by john santu last updated on 07/Apr/20
hihihi.....
hihihi..
Commented by mr W last updated on 07/Apr/20
but for example we don′t have x term,  x^3  term etc.
butforexamplewedonthavexterm,x3termetc.
Commented by jagoll last updated on 07/Apr/20
Σ_(i+j+k ≤11) ^(11)  (((  11)),((i,j,k)) )  (x^5 )^i (x^2 )^j (1)^k   = Σ_(i+j+k ≤ 11) ^(11)  (((  11)),((  i,j,k)) ) x^(5i+2j)  (1)^k   i = 0 ⇒j = 0 ∧k =11,   j = 1 ∧k = 9 , j=2 ∧ k = 7  j = 3 ∧ k = 5 , j = 4 ∧k= 3  j = 5 ∧ k = 1   i=1 ⇒ j = 0 ∧k = 6 , j = 1 ∧k= 4  j = 2 ∧ k = 2 , j = 3 ∧ k = 0  i = 2 ⇒ j = 0∧k=1   Σ terms = 11
11i+j+k11(11i,j,k)(x5)i(x2)j(1)k=11i+j+k11(11i,j,k)x5i+2j(1)ki=0j=0k=11,j=1k=9,j=2k=7j=3k=5,j=4k=3j=5k=1i=1j=0k=6,j=1k=4j=2k=2,j=3k=0i=2j=0k=1Σterms=11
Commented by jagoll last updated on 07/Apr/20
hehehe
hehehe
Commented by ajfour last updated on 07/Apr/20
I still think, mrW Sirs, obvious  answer, correct.
Istillthink,mrWSirs,obviousanswer,correct.
Commented by jagoll last updated on 08/Apr/20
what answer this question sir?
whatanswerthisquestionsir?
Commented by mr W last updated on 08/Apr/20
when expanded, (x^5 +x^2 +1)^(11)  has 51   terms.    (x^5 +x^2 +1)^(11) =Σ_(i+j+k=11) (_(i,j,k) ^(11) )(x^5 )^i (x^2 )^j 1^k   =Σ_(i+j+k=11) (_(i,j,k) ^(11) )x^(5i+2j)   i,j,k≥0  i+j+k=11 ⇒0≤i+j≤11  0≤5i+2j≤55  but from 0 to 55 there are only 51   numbers which is a multiple of 5  (max. 55) or a multiple of 2 (max. 22)  or the sum of both.    following terms don′t exist:  x^1 , x^3 , x^(48) , x^(51) , x^(53)   because you can′t express the numbers  1, 3, 48, 51, 53 as 5i+2j with 0≤i+j≤11.
whenexpanded,(x5+x2+1)11has51terms.(x5+x2+1)11=i+j+k=11(i,j,k11)(x5)i(x2)j1k=i+j+k=11(i,j,k11)x5i+2ji,j,k0i+j+k=110i+j1105i+2j55butfrom0to55thereareonly51numberswhichisamultipleof5(max.55)oramultipleof2(max.22)orthesumofboth.followingtermsdontexist:x1,x3,x48,x51,x53becauseyoucantexpressthenumbers1,3,48,51,53as5i+2jwith0i+j11.

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