Menu Close

Find-coefficient-of-x-29-in-expansion-of-1-x-5-x-7-x-9-1000-




Question Number 161968 by naka3546 last updated on 24/Dec/21
Find  coefficient  of  x^(29)   in  expansion  of   (1+x^5 +x^7 +x^9 )^(1000)  .
Findcoefficientofx29inexpansionof(1+x5+x7+x9)1000.
Answered by mr W last updated on 25/Dec/21
(1+x^5 +x^7 +x^9 )^(1000)   =Σ_(a+b+c+d=1000)  (((1000)),((a,b,c,d)) ) 1^a (x^5 )^b (x^7 )^c (x^9 )^d   =Σ_(a+b+c+d=1000)  (((1000)),((a,b,c,d)) ) x^(5b+7c+9d)   with  (((1000)),((a,b,c,d)) ) =((1000!)/(a!b!c!d!))    for term x^(29) : 5b+7c+9d=29  d=0, c=2, b=3, a=995  d=1, c=0, b=4, a=995  coef. of x^(29)  is therefore   (((1000)),((0,2,3,995)) ) + (((1000)),((1,0,4,995)) )  =((1000!)/(995!))((1/(0!2!3!))+(1/(1!0!4!)))  =((1000!)/(995!))×((1/(12))+(1/(24)))  =((1000!)/(8×995!))  =(1/8)×1000×999×998×997×996  =123 754 368 753 000
(1+x5+x7+x9)1000=a+b+c+d=1000(1000a,b,c,d)1a(x5)b(x7)c(x9)d=a+b+c+d=1000(1000a,b,c,d)x5b+7c+9dwith(1000a,b,c,d)=1000!a!b!c!d!fortermx29:5b+7c+9d=29d=0,c=2,b=3,a=995d=1,c=0,b=4,a=995coef.ofx29istherefore(10000,2,3,995)+(10001,0,4,995)=1000!995!(10!2!3!+11!0!4!)=1000!995!×(112+124)=1000!8×995!=18×1000×999×998×997×996=123754368753000
Commented by naka3546 last updated on 25/Dec/21
Thank  you,  sir.
Thankyou,sir.
Commented by Tawa11 last updated on 25/Dec/21
Great sir
Greatsir
Commented by Jamshidbek last updated on 25/Dec/21
Which is this theorem?
Whichisthistheorem?
Commented by mr W last updated on 25/Dec/21
binomial theorem
binomialtheorem
Commented by mr W last updated on 25/Dec/21
Commented by Jamshidbek last updated on 25/Dec/21
thank you
thankyou

Leave a Reply

Your email address will not be published. Required fields are marked *