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How do you calculate the coefficients of a power series such as this one: [ 1/4x^4 + 1/4x^5 + 1/4x^6]^10. Just the first 3 coefficients, with steps, would be great. I would greatly appreciate any help. Thanks a million and have a great day.

 16.05.2016
 #1
avatar+14538 
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Hello !

 

[ 1/4x^4 + 1/4x^5 + 1/4x^6]^10

 

What do you mean ?

Please ask  Melody in the  English Forum for a result.

 

[ 1/4x^4 + 1/4x^5 + 1/4x^6]^10  =  (x4+x5+x6)101048576

 

or    [ 1/4x^4 + 1/4x^5 + 1/4x^6]^10 = 0         =>   x = 0      and   x=31

 

Gruß radix smiley !

 17.05.2016
 #2
avatar+26397 
0

How do you calculate the coefficients of a power series such as this one:
[ 1/4x^4 + 1/4x^5 + 1/4x^6]^10. Just the first 3 coefficients, with steps, would be great.
I would greatly appreciate any help. Thanks a million and have a great day.

 

[ 14x4+14x5+14x6 ]10=[ 14(x4+x5+x6) ]10=(14)10(x4+x5+x6)10=(1410)(x4+x5+x6)10=(11048576)(x4+x5+x6)10(a+b+c)n=nk=0ki=0(nk)(ki)ankbkici|(nk)+(ki)+(i)=n(x4+x5+x6)10=10k=0ki=0(10k)(ki)(x4)10k(x5)ki(x6)i|(10k)+(ki)+(i)=10

 

(k=0,i=0):(100)(00)(x4)10(x5)0(x6)0=x40(k=1,i=0):(101)(10)(x4)9(x5)1(x6)0=10x41(k=1,i=1):(101)(11)(x4)9(x5)0(x6)1=10x42(k=2,i=0):(102)(20)(x4)8(x5)2(x6)0=45x42(k=2,i=1):(102)(21)(x4)8(x5)1(x6)1=90x43(k=2,i=2):(102)(22)(x4)8(x5)0(x6)2=45x44(k=3,i=0):(103)(30)(x4)7(x5)3(x6)0=120x43(k=3,i=1):(103)(31)(x4)7(x5)2(x6)1=360x44(k=3,i=2):(103)(32)(x4)7(x5)1(x6)2=360x45(k=3,i=3):(103)(33)(x4)7(x5)0(x6)3=120x46

 

x4010x4110x4245x4290x4345x44120x43360x44360x45120x46sumx4010x4155x42210x43

 

[ 14x4+14x5+14x6 ]10=11048576x40+101048576x41+551048576x42+2101048576x43+6151048576x44+14521048576x45+28501048576x46+47401048576x47+67651048576x48+83501048576x49+89531048576x50+83501048576x51+67651048576x52+47401048576x53+28501048576x54+14521048576x55+6151048576x56+2101048576x57+551048576x58+101048576x59+11048576x60

heureka  17.05.2016

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