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.
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=−3√−1
Gruß radix !
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.
[ 14⋅x4+14⋅x5+14⋅x6 ]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=n∑k=0k∑i=0(nk)(ki)an−kbk−ici|(n−k)+(k−i)+(i)=n(x4+x5+x6)10=10∑k=0k∑i=0(10k)(ki)(x4)10−k(x5)k−i(x6)i|(10−k)+(k−i)+(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=10⋅x41(k=1,i=1):(101)(11)(x4)9(x5)0(x6)1=10⋅x42(k=2,i=0):(102)(20)(x4)8(x5)2(x6)0=45⋅x42(k=2,i=1):(102)(21)(x4)8(x5)1(x6)1=90⋅x43(k=2,i=2):(102)(22)(x4)8(x5)0(x6)2=45⋅x44(k=3,i=0):(103)(30)(x4)7(x5)3(x6)0=120⋅x43(k=3,i=1):(103)(31)(x4)7(x5)2(x6)1=360⋅x44(k=3,i=2):(103)(32)(x4)7(x5)1(x6)2=360⋅x45(k=3,i=3):(103)(33)(x4)7(x5)0(x6)3=120⋅x46⋯
x4010⋅x4110⋅x4245⋅x4290⋅x4345⋅x44120⋅x43360⋅x44360⋅x45120⋅x46⋯⋯⋯sumx4010⋅x4155⋅x42210⋅x43⋯⋯⋯
[ 14⋅x4+14⋅x5+14⋅x6 ]10=11048576⋅x40+101048576⋅x41+551048576⋅x42+2101048576⋅x43+6151048576⋅x44+14521048576⋅x45+28501048576⋅x46+47401048576⋅x47+67651048576⋅x48+83501048576⋅x49+89531048576⋅x50+83501048576⋅x51+67651048576⋅x52+47401048576⋅x53+28501048576⋅x54+14521048576⋅x55+6151048576⋅x56+2101048576⋅x57+551048576⋅x58+101048576⋅x59+11048576⋅x60