Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 82149 by TawaTawa last updated on 18/Feb/20

Evaluate the greatest coefficient of (7 − 5x)^(− 3)

$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{of}\:\:\:\:\left(\mathrm{7}\:−\:\mathrm{5x}\right)^{−\:\mathrm{3}} \\ $$

Commented by TawaTawa last updated on 18/Feb/20

Please working sir. I appreciate.

$$\mathrm{Please}\:\mathrm{working}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 18/Feb/20

(7−5x)^(−3) =7^(−3) (1−((5x)/7))^(−3) =7^(−3) Σ_(k=0) ^∞ C_2 ^(k+2) (((5x)/7))^k =Σ_(k=0) ^∞ (5^k /7^(k+3) )C_2 ^(k+2) x^k a_k =(5^k /7^(k+3) )C_2 ^(k+2) if a_n is the greatest, then a_n ≥a_(n+1) , i.e. (5^n /7^(n+3) )C_2 ^(n+2) ≥(5^(n+1) /7^(n+4) )C_2 ^(n+3) C_2 ^(n+2) ≥(5/7)C_2 ^(n+3) (((n+2)(n+1))/(2!))≥(5/7)×(((n+3)(n+2))/(2!)) 7(n+1)≥5(n+3) n≥4 that means term a_4 is the greatest. a_4 =(5^4 /7^7 )C_2 ^6 =((5^4 ×6×5)/(7^7 ×2!))=((3×5^5 )/7^7 )=((9375)/(823543))

$$\left(\mathrm{7}−\mathrm{5}{x}\right)^{−\mathrm{3}} \\ $$$$=\mathrm{7}^{−\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{5}{x}}{\mathrm{7}}\right)^{−\mathrm{3}} \\ $$$$=\mathrm{7}^{−\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} \left(\frac{\mathrm{5}{x}}{\mathrm{7}}\right)^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{5}^{{k}} }{\mathrm{7}^{{k}+\mathrm{3}} }{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${a}_{{k}} =\frac{\mathrm{5}^{{k}} }{\mathrm{7}^{{k}+\mathrm{3}} }{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} \\ $$$${if}\:{a}_{{n}} \:{is}\:{the}\:{greatest},\:{then}\:{a}_{{n}} \geqslant{a}_{{n}+\mathrm{1}} ,\:{i}.{e}. \\ $$$$\frac{\mathrm{5}^{{n}} }{\mathrm{7}^{{n}+\mathrm{3}} }{C}_{\mathrm{2}} ^{{n}+\mathrm{2}} \geqslant\frac{\mathrm{5}^{{n}+\mathrm{1}} }{\mathrm{7}^{{n}+\mathrm{4}} }{C}_{\mathrm{2}} ^{{n}+\mathrm{3}} \\ $$$${C}_{\mathrm{2}} ^{{n}+\mathrm{2}} \geqslant\frac{\mathrm{5}}{\mathrm{7}}{C}_{\mathrm{2}} ^{{n}+\mathrm{3}} \\ $$$$\frac{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{\mathrm{2}!}\geqslant\frac{\mathrm{5}}{\mathrm{7}}×\frac{\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}!} \\ $$$$\mathrm{7}\left({n}+\mathrm{1}\right)\geqslant\mathrm{5}\left({n}+\mathrm{3}\right) \\ $$$${n}\geqslant\mathrm{4} \\ $$$${that}\:{means}\:{term}\:{a}_{\mathrm{4}} \:{is}\:{the}\:{greatest}. \\ $$$${a}_{\mathrm{4}} =\frac{\mathrm{5}^{\mathrm{4}} }{\mathrm{7}^{\mathrm{7}} }{C}_{\mathrm{2}} ^{\mathrm{6}} =\frac{\mathrm{5}^{\mathrm{4}} ×\mathrm{6}×\mathrm{5}}{\mathrm{7}^{\mathrm{7}} ×\mathrm{2}!}=\frac{\mathrm{3}×\mathrm{5}^{\mathrm{5}} }{\mathrm{7}^{\mathrm{7}} }=\frac{\mathrm{9375}}{\mathrm{823543}} \\ $$

Commented by TawaTawa last updated on 18/Feb/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 19/Feb/20

is my working clear and understood?

$${is}\:{my}\:{working}\:{clear}\:{and}\:{understood}? \\ $$

Commented by TawaTawa last updated on 19/Feb/20

It is only the third line am finding difficult to understand sir

$$\mathrm{It}\:\mathrm{is}\:\mathrm{only}\:\mathrm{the}\:\mathrm{third}\:\mathrm{line}\:\mathrm{am}\:\mathrm{finding}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{sir} \\ $$

Commented by TawaTawa last updated on 19/Feb/20

How to get the combination

$$\mathrm{How}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{combination} \\ $$

Commented by mr W last updated on 19/Feb/20

you must have learnt (a+b)^n . if not, try to study about “binomial theorem”. we have here a=1, b=−((5x)/7) and n=−3.

$${you}\:{must}\:{have}\:{learnt}\:\left({a}+{b}\right)^{{n}} .\:{if}\:{not}, \\ $$$${try}\:{to}\:{study}\:{about}\:``{binomial}\:{theorem}''. \\ $$$${we}\:{have}\:{here}\:{a}=\mathrm{1},\:{b}=−\frac{\mathrm{5}{x}}{\mathrm{7}}\:{and}\:{n}=−\mathrm{3}. \\ $$

Commented by TawaTawa last updated on 19/Feb/20

Yes sir. I know (a + b)^n How can i get: ^(k + 2) C_2 (((5x)/7))^k Thanks for your help sir

$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{know}\:\:\left(\mathrm{a}\:+\:\mathrm{b}\right)^{\mathrm{n}} \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}:\:\:\:\:\:\overset{\mathrm{k}\:+\:\mathrm{2}} {\:}\mathrm{C}_{\mathrm{2}} \left(\frac{\mathrm{5x}}{\mathrm{7}}\right)^{\mathrm{k}} \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 19/Feb/20

Commented by TawaTawa last updated on 19/Feb/20

God bless you sir. I understand now. Thanks for your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com