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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) .

$${Find}\:\:{coefficient}\:\:{of}\:\:{x}^{\mathrm{29}} \:\:{in}\:\:{expansion}\:\:{of}\:\:\:\left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{7}} +{x}^{\mathrm{9}} \right)^{\mathrm{1000}} \:. \\ $$

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

$$\left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{7}} +{x}^{\mathrm{9}} \right)^{\mathrm{1000}} \\ $$$$=\underset{{a}+{b}+{c}+{d}=\mathrm{1000}} {\sum}\begin{pmatrix}{\mathrm{1000}}\\{{a},{b},{c},{d}}\end{pmatrix}\:\mathrm{1}^{{a}} \left({x}^{\mathrm{5}} \right)^{{b}} \left({x}^{\mathrm{7}} \right)^{{c}} \left({x}^{\mathrm{9}} \right)^{{d}} \\ $$$$=\underset{{a}+{b}+{c}+{d}=\mathrm{1000}} {\sum}\begin{pmatrix}{\mathrm{1000}}\\{{a},{b},{c},{d}}\end{pmatrix}\:{x}^{\mathrm{5}{b}+\mathrm{7}{c}+\mathrm{9}{d}} \\ $$$${with}\:\begin{pmatrix}{\mathrm{1000}}\\{{a},{b},{c},{d}}\end{pmatrix}\:=\frac{\mathrm{1000}!}{{a}!{b}!{c}!{d}!} \\ $$$$ \\ $$$${for}\:{term}\:{x}^{\mathrm{29}} :\:\mathrm{5}{b}+\mathrm{7}{c}+\mathrm{9}{d}=\mathrm{29} \\ $$$${d}=\mathrm{0},\:{c}=\mathrm{2},\:{b}=\mathrm{3},\:{a}=\mathrm{995} \\ $$$${d}=\mathrm{1},\:{c}=\mathrm{0},\:{b}=\mathrm{4},\:{a}=\mathrm{995} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{29}} \:{is}\:{therefore} \\ $$$$\begin{pmatrix}{\mathrm{1000}}\\{\mathrm{0},\mathrm{2},\mathrm{3},\mathrm{995}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{1000}}\\{\mathrm{1},\mathrm{0},\mathrm{4},\mathrm{995}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1000}!}{\mathrm{995}!}\left(\frac{\mathrm{1}}{\mathrm{0}!\mathrm{2}!\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{1}!\mathrm{0}!\mathrm{4}!}\right) \\ $$$$=\frac{\mathrm{1000}!}{\mathrm{995}!}×\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{24}}\right) \\ $$$$=\frac{\mathrm{1000}!}{\mathrm{8}×\mathrm{995}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}×\mathrm{1000}×\mathrm{999}×\mathrm{998}×\mathrm{997}×\mathrm{996} \\ $$$$=\mathrm{123}\:\mathrm{754}\:\mathrm{368}\:\mathrm{753}\:\mathrm{000} \\ $$

Commented by naka3546 last updated on 25/Dec/21

Thank you, sir.

$${Thank}\:\:{you},\:\:{sir}. \\ $$

Commented by Tawa11 last updated on 25/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Jamshidbek last updated on 25/Dec/21

Which is this theorem?

$$\mathrm{Which}\:\mathrm{is}\:\mathrm{this}\:\mathrm{theorem}? \\ $$

Commented by mr W last updated on 25/Dec/21

binomial theorem

$${binomial}\:{theorem} \\ $$

Commented by mr W last updated on 25/Dec/21

Commented by Jamshidbek last updated on 25/Dec/21

thank you

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

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