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Question Number 8217 by tawakalitu last updated on 02/Oct/16
what is the coefficient of x^3  in the expansion  of (1 + x + x^2  + x^3  + x^4  + x^5 )^6
whatisthecoefficientofx3intheexpansionof(1+x+x2+x3+x4+x5)6
Commented by 123456 last updated on 04/Oct/16
1∙2∙3∙4∙5∙6  1+1+1=1+2=3  3:6  1+2:6∙5=30  1+1+1:((6∙5∙4)/(3!))=((6∙5∙4)/(3∙2))=5∙4=20  6+30+20=56
1234561+1+1=1+2=33:61+2:65=301+1+1:6543!=65432=54=206+30+20=56
Commented by sandy_suhendra last updated on 04/Oct/16
this is so simple, but can you more explain it?
thisissosimple,butcanyoumoreexplainit?
Commented by 123456 last updated on 04/Oct/16
we have  (1+x+x^2 +x^3 +x^4 +x^5 )^6 =  (1+∙∙∙+x^5 )...(1+∙∙∙+x^5 )  recal x^3 =x^2 x=xxx  we have to pick one factor from each  (), if we pick x^3 , all other will be 1  we have 6 () so 6 ways  if we pick x^2 , we need to pick a x and  all other will be 1, so  6() × 5() = 30 ways  if we pick x, we pick x^2  (above case)  or x and x, so  6 () × 5() × 4() =“120 ways”  however  x×x×x=x×x×x (3!=6 ways of perm x)  so  ((120)/6)=20 ways  since all coef are 1 the coef is the sum  of number of way  6+30+20=56
wehave(1+x+x2+x3+x4+x5)6=(1++x5)(1++x5)recalx3=x2x=xxxwehavetopickonefactorfromeach(),ifwepickx3,allotherwillbe1wehave6()so6waysifwepickx2,weneedtopickaxandallotherwillbe1,so6()×5()=30waysifwepickx,wepickx2(abovecase)orxandx,so6()×5()×4()=120wayshoweverx×x×x=x×x×x(3!=6waysofpermx)so1206=20wayssinceallcoefare1thecoefisthesumofnumberofway6+30+20=56
Commented by sandy_suhendra last updated on 04/Oct/16
great! thank′s for your answer
great!thanksforyouranswer
Commented by tawakalitu last updated on 04/Oct/16
Thanks so much. i really appreciate.
Thankssomuch.ireallyappreciate.
Answered by sandy_suhendra last updated on 03/Oct/16
[(1+x+x^2 )+x^3 (1+x+x^2 )]^6   =[(x^3 +1)(x^2 +x+1)]^6   =(x^3 +1)^6 (x^2 +x+1)^6   (x^3 +1)^6 =(x^3 )^6 +6(x^3 )^5 +...+6x^3 +1  [x^2 +(x+1)]^6   =1(x^2 )^6 +6(x^2 )^5 (x+1)+...+6x^2 (x+1)^5 +1(x+1)^6   6x^2 (x+1)^5 =6x^2 (x^5 +5x^4 +...+5x+1)⇒30x^3   (x+1)^6 =x^6 +6x^5 +...+20x^3 +...+1  the term of x^3  comes from ⇒ 6x^3 +30x^3 +20x^3 =56x^3   so the coefficient of x^3  is 56
[(1+x+x2)+x3(1+x+x2)]6=[(x3+1)(x2+x+1)]6=(x3+1)6(x2+x+1)6(x3+1)6=(x3)6+6(x3)5++6x3+1[x2+(x+1)]6=1(x2)6+6(x2)5(x+1)++6x2(x+1)5+1(x+1)66x2(x+1)5=6x2(x5+5x4++5x+1)30x3(x+1)6=x6+6x5++20x3++1thetermofx3comesfrom6x3+30x3+20x3=56x3sothecoefficientofx3is56
Commented by tawakalitu last updated on 03/Oct/16
Thanks so much.
Thankssomuch.

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