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Question Number 84998 by jagoll last updated on 18/Mar/20
what is coefficient of x^(29) in expression (1+x^5 +x^7 +x^9 )^(29)
$$\mathrm{what}\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{29}} \\ $$$$\mathrm{in}\:\mathrm{expression}\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{7}} +\mathrm{x}^{\mathrm{9}} \right)^{\mathrm{29}} \\ $$
Answered by mind is power last updated on 18/Mar/20
29=5a+7b+9c a+b+c+d=29 c=1⇒20=5a+7b⇒b=0,a=4 ⇒d=24 c=0 29=5a+7b⇒a=3,b=2, d=24 k=(a,b,c) (1+x^5 +x^7 +x^9 )^(29) =Σ_(a+b+c=29−d) ((29!)/(a!.b!.c!))x^(5a+7b+9c) =(((29!)/(4!.24!))+((29!)/(3!.2!.24!)))x^(29)
$$\mathrm{29}=\mathrm{5}{a}+\mathrm{7}{b}+\mathrm{9}{c} \\ $$$${a}+{b}+{c}+{d}=\mathrm{29} \\ $$$${c}=\mathrm{1}\Rightarrow\mathrm{20}=\mathrm{5}{a}+\mathrm{7}{b}\Rightarrow{b}=\mathrm{0},{a}=\mathrm{4} \\ $$$$\Rightarrow{d}=\mathrm{24} \\ $$$${c}=\mathrm{0} \\ $$$$\mathrm{29}=\mathrm{5}{a}+\mathrm{7}{b}\Rightarrow{a}=\mathrm{3},{b}=\mathrm{2}, \\ $$$${d}=\mathrm{24} \\ $$$${k}=\left({a},{b},{c}\right) \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{7}} +{x}^{\mathrm{9}} \right)^{\mathrm{29}} =\underset{{a}+{b}+{c}=\mathrm{29}−{d}} {\sum}\frac{\mathrm{29}!}{{a}!.{b}!.{c}!}{x}^{\mathrm{5}{a}+\mathrm{7}{b}+\mathrm{9}{c}} \\ $$$$=\left(\frac{\mathrm{29}!}{\mathrm{4}!.\mathrm{24}!}+\frac{\mathrm{29}!}{\mathrm{3}!.\mathrm{2}!.\mathrm{24}!}\right){x}^{\mathrm{29}} \\ $$
Commented by jagoll last updated on 18/Mar/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 18/Mar/20
yes sir correct. via wolframalpha
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{correct}.\:\mathrm{via}\:\mathrm{wolframalpha} \\ $$
Commented by mind is power last updated on 18/Mar/20
yeah i forget 24!
$${yeah}\:{i}\:{forget}\:\mathrm{24}! \\ $$$$ \\ $$
Answered by mr W last updated on 18/Mar/20
(1+x^5 +x^7 +x^9 )^(29) =Σ_(a+b+c+d=29) (_(a,b,c,d) ^(29) )1^d (x^5 )^a (x^7 )^b (x^9 )^c =Σ_(a+b+c+d=29) (_(a,b,c,d) ^(29) )x^(5a+7b+9c) with (_(a,b,c,d) ^(29) )=((29!)/(a!b!c!d!)) example: term x^(29) : 5a+7b+9c=29 ⇒a=3, b=2, c=0 ⇒d=24 ⇒a=4, b=0, c=1 ⇒d=24 the coefficient of x^(29) : ((29!)/(3!2!24!))+((29!)/(4!24!))=29×28×27×26×25((1/(12))+(1/(24))) =1781325 example: term x^(50) : 5a+7b+9c=50 ⇒a=3, b=5, c=0 ⇒d=21 ⇒a=10, b=0, c=0 ⇒d=19 ⇒a=1, b=0, c=5 ⇒d=23 the coefficient of x^(50) : ((29!)/(3!5!21!))+((29!)/(10!19!))+((29!)/(5!23!))=263 240 250
$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{7}} +\mathrm{x}^{\mathrm{9}} \right)^{\mathrm{29}} \\ $$$$=\underset{{a}+{b}+{c}+{d}=\mathrm{29}} {\sum}\left(_{{a},{b},{c},{d}} ^{\mathrm{29}} \right)\mathrm{1}^{{d}} \left({x}^{\mathrm{5}} \right)^{{a}} \left({x}^{\mathrm{7}} \right)^{{b}} \left({x}^{\mathrm{9}} \right)^{{c}} \\ $$$$=\underset{{a}+{b}+{c}+{d}=\mathrm{29}} {\sum}\left(_{{a},{b},{c},{d}} ^{\mathrm{29}} \right){x}^{\mathrm{5}{a}+\mathrm{7}{b}+\mathrm{9}{c}} \\ $$$${with}\:\left(_{{a},{b},{c},{d}} ^{\mathrm{29}} \right)=\frac{\mathrm{29}!}{{a}!{b}!{c}!{d}!} \\ $$$$ \\ $$$${example}:\:{term}\:{x}^{\mathrm{29}} : \\ $$$$\mathrm{5}{a}+\mathrm{7}{b}+\mathrm{9}{c}=\mathrm{29} \\ $$$$\Rightarrow{a}=\mathrm{3},\:{b}=\mathrm{2},\:{c}=\mathrm{0}\:\Rightarrow{d}=\mathrm{24} \\ $$$$\Rightarrow{a}=\mathrm{4},\:{b}=\mathrm{0},\:{c}=\mathrm{1}\:\Rightarrow{d}=\mathrm{24} \\ $$$${the}\:{coefficient}\:{of}\:{x}^{\mathrm{29}} : \\ $$$$\frac{\mathrm{29}!}{\mathrm{3}!\mathrm{2}!\mathrm{24}!}+\frac{\mathrm{29}!}{\mathrm{4}!\mathrm{24}!}=\mathrm{29}×\mathrm{28}×\mathrm{27}×\mathrm{26}×\mathrm{25}\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{24}}\right) \\ $$$$=\mathrm{1781325} \\ $$$$ \\ $$$${example}:\:{term}\:{x}^{\mathrm{50}} : \\ $$$$\mathrm{5}{a}+\mathrm{7}{b}+\mathrm{9}{c}=\mathrm{50} \\ $$$$\Rightarrow{a}=\mathrm{3},\:{b}=\mathrm{5},\:{c}=\mathrm{0}\:\Rightarrow{d}=\mathrm{21} \\ $$$$\Rightarrow{a}=\mathrm{10},\:{b}=\mathrm{0},\:{c}=\mathrm{0}\:\Rightarrow{d}=\mathrm{19} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=\mathrm{0},\:{c}=\mathrm{5}\:\Rightarrow{d}=\mathrm{23} \\ $$$${the}\:{coefficient}\:{of}\:{x}^{\mathrm{50}} : \\ $$$$\frac{\mathrm{29}!}{\mathrm{3}!\mathrm{5}!\mathrm{21}!}+\frac{\mathrm{29}!}{\mathrm{10}!\mathrm{19}!}+\frac{\mathrm{29}!}{\mathrm{5}!\mathrm{23}!}=\mathrm{263}\:\mathrm{240}\:\mathrm{250} \\ $$
Commented by mr W last updated on 18/Mar/20
Commented by mr W last updated on 18/Mar/20
my result coincides with that from wolframalpha, see below.
$${my}\:{result}\:{coincides}\:{with}\:{that}\:{from} \\ $$$${wolframalpha},\:{see}\:{below}. \\ $$
Commented by jagoll last updated on 19/Mar/20
sir w . thank you. i′m wrong operate my wolframaplha.
$$\mathrm{sir}\:\mathrm{w}\:.\:\mathrm{thank}\:\mathrm{you}.\:\mathrm{i}'\mathrm{m}\:\mathrm{wrong} \\ $$$$\mathrm{operate}\:\mathrm{my}\:\mathrm{wolframaplha}.\: \\ $$

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