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Question Number 102278 by I want to learn more last updated on 08/Jul/20

Find the greatest coefficient in the following without actually expand. (i) (5 − 3x)^(10) (ii) (5 + 3x)^(− 10)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{without} \\ $$$$\mathrm{actually}\:\mathrm{expand}. \\ $$$$\left(\mathrm{i}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:−\:\:\mathrm{3x}\right)^{\mathrm{10}} \\ $$$$\left(\mathrm{ii}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:+\:\:\mathrm{3x}\right)^{−\:\mathrm{10}} \\ $$

Commented by mr W last updated on 08/Jul/20

see Q94397

$${see}\:{Q}\mathrm{94397} \\ $$

Commented by mr W last updated on 08/Jul/20

see Q91464

$${see}\:{Q}\mathrm{91464} \\ $$

Commented by I want to learn more last updated on 08/Jul/20

Sir am going say 5^(10) Σ_(k = 0) ^(10) ^(10) C_k (− ((3x)/5))^k Sir am going say 5^(10) Σ_(k = 0) ^(10) ^(10) C_k (− (3/5))^k .x^k Am concern about the minus. And here, (3 + 2x)^(− 7) Sir am going say (1/(3^7 )) Σ_(k = 0) ^∞ ^(k + 6) C_k (− ((2x)/3))^k The signs is my concern. Just like the opposite sign to the one you solved sir.

$$\mathrm{Sir}\:\mathrm{am}\:\mathrm{going}\:\mathrm{say}\:\:\:\:\mathrm{5}^{\mathrm{10}} \underset{\mathrm{k}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{k}} \:\left(−\:\frac{\mathrm{3x}}{\mathrm{5}}\right)^{\mathrm{k}} \\ $$$$\mathrm{Sir}\:\mathrm{am}\:\mathrm{going}\:\mathrm{say}\:\:\:\:\mathrm{5}^{\mathrm{10}} \underset{\mathrm{k}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{k}} \:\left(−\:\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{k}} .\mathrm{x}^{\mathrm{k}} \\ $$$$\mathrm{Am}\:\mathrm{concern}\:\mathrm{about}\:\mathrm{the}\:\mathrm{minus}. \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{here}, \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{3}\:\:+\:\:\mathrm{2x}\right)^{−\:\mathrm{7}} \: \\ $$$$\mathrm{Sir}\:\mathrm{am}\:\mathrm{going}\:\mathrm{say}\:\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{7}} \:}\:\underset{\mathrm{k}\:\:=\:\:\mathrm{0}} {\overset{\infty} {\sum}}\:\overset{\mathrm{k}\:\:+\:\:\mathrm{6}} {\:}\mathrm{C}_{\mathrm{k}} \:\left(−\:\frac{\mathrm{2x}}{\mathrm{3}}\right)^{\mathrm{k}} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{signs}\:\mathrm{is}\:\mathrm{my}\:\mathrm{concern}. \\ $$$$\mathrm{Just}\:\mathrm{like}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sign}\:\mathrm{to}\:\mathrm{the}\:\mathrm{one}\:\mathrm{you}\:\mathrm{solved}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 08/Jul/20

i showed you two different ways how to treat the “−” sign.

$${i}\:{showed}\:{you}\:{two}\:{different}\:{ways}\:{how} \\ $$$${to}\:{treat}\:{the}\:``−''\:{sign}. \\ $$

Answered by john santu last updated on 08/Jul/20

(1)C_5 ^(10) 5^5 (−3x)^6 = ((10.9.8.6)/(5.4.3.2.1)). 5^5 .(−3)^6 = 5^5 . 72. 3^6

$$\left(\mathrm{1}\right){C}_{\mathrm{5}} ^{\mathrm{10}} \:\mathrm{5}^{\mathrm{5}} \:\left(−\mathrm{3}{x}\right)^{\mathrm{6}} \:=\:\frac{\mathrm{10}.\mathrm{9}.\mathrm{8}.\mathrm{6}}{\mathrm{5}.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{1}}.\:\mathrm{5}^{\mathrm{5}} .\left(−\mathrm{3}\right)^{\mathrm{6}} \\ $$$$=\:\mathrm{5}^{\mathrm{5}} .\:\mathrm{72}.\:\mathrm{3}^{\mathrm{6}} \: \\ $$

Commented by I want to learn more last updated on 08/Jul/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 08/Jul/20

(i) (5−3x)^(10) =Σ_(k=0) ^(10) C_k ^(10) 5^(10−k) (−3x)^k =Σ_(k=0) ^(10) (−1)^k C_k ^(10) 3^k 5^(10−k) x^k a_k =(−1)^k C_k ^(10) 3^k 5^(10−k) for greatest coefficient a_k we should only look at the positive values, i.e. if k is even: k=2n a_(2n) =C_(2n) ^(10) 3^(2n) 5^(10−2n) =5^(10) C_(2n) ^(10) ((9/(25)))^n let C_(2n) ^(10) ((9/(25)))^n >C_(2(n+1)) ^(10) ((9/(25)))^(n+1) ((10!)/((2n)!(10−2n)!))((9/(25)))^n >((10!)/((2n+2)!(10−2n−2)!))((9/(25)))^(n+1) (1/((10−2n)(10−2n−1)))>(1/((2n+2)(2n+1)))×(9/(25)) ((25)/(45−19n+2n^2 ))>(9/(2n^2 +3n+1)) 16n^2 +123n−190>0 n>((−123+(√(123^2 +4×16×190)))/(32))≈1.3 ⇒n=2 i.e. a_4 is the greates coefficient. a_4 =5^(10) ×C_4 ^(10) ×((9/(25)))^2 =265 781 250

$$\left({i}\right) \\ $$$$\left(\mathrm{5}−\mathrm{3}{x}\right)^{\mathrm{10}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}{C}_{{k}} ^{\mathrm{10}} \mathrm{5}^{\mathrm{10}−{k}} \left(−\mathrm{3}{x}\right)^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{\mathrm{10}} \mathrm{3}^{{k}} \mathrm{5}^{\mathrm{10}−{k}} {x}^{{k}} \\ $$$${a}_{{k}} =\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{\mathrm{10}} \mathrm{3}^{{k}} \mathrm{5}^{\mathrm{10}−{k}} \\ $$$${for}\:{greatest}\:{coefficient}\:{a}_{{k}} \:{we}\:{should} \\ $$$${only}\:{look}\:{at}\:{the}\:{positive}\:{values},\:{i}.{e}. \\ $$$${if}\:{k}\:{is}\:{even}:\:{k}=\mathrm{2}{n} \\ $$$${a}_{\mathrm{2}{n}} ={C}_{\mathrm{2}{n}} ^{\mathrm{10}} \mathrm{3}^{\mathrm{2}{n}} \mathrm{5}^{\mathrm{10}−\mathrm{2}{n}} \\ $$$$=\mathrm{5}^{\mathrm{10}} {C}_{\mathrm{2}{n}} ^{\mathrm{10}} \left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{{n}} \\ $$$${let}\:{C}_{\mathrm{2}{n}} ^{\mathrm{10}} \left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{{n}} >{C}_{\mathrm{2}\left({n}+\mathrm{1}\right)} ^{\mathrm{10}} \left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{10}!}{\left(\mathrm{2}{n}\right)!\left(\mathrm{10}−\mathrm{2}{n}\right)!}\left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{{n}} >\frac{\mathrm{10}!}{\left(\mathrm{2}{n}+\mathrm{2}\right)!\left(\mathrm{10}−\mathrm{2}{n}−\mathrm{2}\right)!}\left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{10}−\mathrm{2}{n}\right)\left(\mathrm{10}−\mathrm{2}{n}−\mathrm{1}\right)}>\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}×\frac{\mathrm{9}}{\mathrm{25}} \\ $$$$\frac{\mathrm{25}}{\mathrm{45}−\mathrm{19}{n}+\mathrm{2}{n}^{\mathrm{2}} }>\frac{\mathrm{9}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}} \\ $$$$\mathrm{16}{n}^{\mathrm{2}} +\mathrm{123}{n}−\mathrm{190}>\mathrm{0} \\ $$$${n}>\frac{−\mathrm{123}+\sqrt{\mathrm{123}^{\mathrm{2}} +\mathrm{4}×\mathrm{16}×\mathrm{190}}}{\mathrm{32}}\approx\mathrm{1}.\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$${i}.{e}.\:{a}_{\mathrm{4}} \:{is}\:{the}\:{greates}\:{coefficient}. \\ $$$${a}_{\mathrm{4}} =\mathrm{5}^{\mathrm{10}} ×{C}_{\mathrm{4}} ^{\mathrm{10}} ×\left(\frac{\mathrm{9}}{\mathrm{25}}\right)^{\mathrm{2}} =\mathrm{265}\:\mathrm{781}\:\mathrm{250} \\ $$

Commented by I want to learn more last updated on 09/Jul/20

I really appreciate sir

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 08/Jul/20

(ii) (5+3x)^(−10) =(1/5^(10) )(1+((3x)/5))^(−10) =(1/5^(10) )Σ_(k=0) ^∞ C_9 ^(k+9) (−(3/5))^k x^k let C_9 ^(k+9) ((3/5))^k >C_9 ^(k+10) ((3/5))^(k+1) (((k+9)!)/(9!k!))>(((k+10)!)/(9!(k+1)!))×(3/5) 1>(((k+10))/((k+1)))×(3/5) 2k>25 k>((25)/2)=12.5 ⇒k=14 since we search even terms i.e. a_(14) is the greatest coefficient. a_(14) =(1/5^(10) )×C_9 ^(23) ×((3/5))^(14) =((817190×3^(14) )/5^(24) )

$$\left({ii}\right) \\ $$$$\left(\mathrm{5}+\mathrm{3}{x}\right)^{−\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{10}} }\left(\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{5}}\right)^{−\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{10}} }\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{9}} ^{{k}+\mathrm{9}} \left(−\frac{\mathrm{3}}{\mathrm{5}}\right)^{{k}} {x}^{{k}} \\ $$$${let}\:{C}_{\mathrm{9}} ^{{k}+\mathrm{9}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{k}} >{C}_{\mathrm{9}} ^{{k}+\mathrm{10}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{k}+\mathrm{1}} \\ $$$$\frac{\left({k}+\mathrm{9}\right)!}{\mathrm{9}!{k}!}>\frac{\left({k}+\mathrm{10}\right)!}{\mathrm{9}!\left({k}+\mathrm{1}\right)!}×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{1}>\frac{\left({k}+\mathrm{10}\right)}{\left({k}+\mathrm{1}\right)}×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{2}{k}>\mathrm{25} \\ $$$${k}>\frac{\mathrm{25}}{\mathrm{2}}=\mathrm{12}.\mathrm{5} \\ $$$$\Rightarrow{k}=\mathrm{14}\:{since}\:{we}\:{search}\:{even}\:{terms} \\ $$$${i}.{e}.\:{a}_{\mathrm{14}} \:{is}\:{the}\:{greatest}\:{coefficient}. \\ $$$${a}_{\mathrm{14}} =\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{10}} }×{C}_{\mathrm{9}} ^{\mathrm{23}} ×\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{14}} =\frac{\mathrm{817190}×\mathrm{3}^{\mathrm{14}} }{\mathrm{5}^{\mathrm{24}} } \\ $$

Commented by I want to learn more last updated on 09/Jul/20

I really appreciate sir

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

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