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Question Number 77902 by TawaTawa last updated on 12/Jan/20

Commented by mr W last updated on 12/Jan/20

try to solve by yourself. i think you can solve. but the question is not a good one. it says only α,β,γ>0. but α+β−γ could be >0, =0 or <0 which makes the answer not unique. so assume α+β−γ≥0.

$${try}\:{to}\:{solve}\:{by}\:{yourself}.\:{i}\:{think}\:{you} \\ $$$${can}\:{solve}. \\ $$$${but}\:{the}\:{question}\:{is}\:{not}\:{a}\:{good}\:{one}.\:{it}\:{says} \\ $$$${only}\:\alpha,\beta,\gamma>\mathrm{0}.\:{but}\:\alpha+\beta−\gamma\:{could}\:{be}\:>\mathrm{0}, \\ $$$$=\mathrm{0}\:{or}\:<\mathrm{0}\:{which}\:{makes}\:{the}\:{answer}\:{not} \\ $$$${unique}.\:{so}\:{assume}\:\alpha+\beta−\gamma\geqslant\mathrm{0}. \\ $$

Commented by TawaTawa last updated on 12/Jan/20

Sir, i asked questions i cannot solve. I want to know the workings so that i can answer correctly in exam. You can help me if you are chanced sir. Sorry to disturb you.

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{asked}\:\mathrm{questions}\:\mathrm{i}\:\mathrm{cannot}\:\mathrm{solve}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{the}\:\mathrm{workings}\:\mathrm{so}\:\mathrm{that}\:\mathrm{i}\:\mathrm{can}\:\mathrm{answer} \\ $$$$\mathrm{correctly}\:\mathrm{in}\:\mathrm{exam}. \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{help}\:\mathrm{me}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chanced}\:\mathrm{sir}. \\ $$$$\mathrm{Sorry}\:\mathrm{to}\:\mathrm{disturb}\:\mathrm{you}. \\ $$

Commented by mr W last updated on 12/Jan/20

S=sum of all coef. of (ax+αy−γz)^n ⇒S=(α+β−γ)^n ⇒S^(1/n) =α+β−γ ⇒{S^(1/n) +1}^n =(α+β−γ+1)^n (S/({S^(1/n) +1}^n ))=(((α+β−γ)^n )/((α+β−γ+1)^n ))=(((α+β−γ)/(α+β−γ+1)))^n =(1−(1/(α+β−γ+1)))^n lim_(n→∞) (S/({S^(1/n) +1}^n ))=lim_(n→∞) (1−(1/(α+β−γ+1)))^n =0, assume α+β−γ≥0 ⇒answer (d)

$${S}={sum}\:{of}\:{all}\:{coef}.\:{of}\:\left({ax}+\alpha{y}−\gamma{z}\right)^{{n}} \\ $$$$\Rightarrow{S}=\left(\alpha+\beta−\gamma\right)^{{n}} \\ $$$$\Rightarrow{S}^{\frac{\mathrm{1}}{{n}}} =\alpha+\beta−\gamma \\ $$$$\Rightarrow\left\{{S}^{\frac{\mathrm{1}}{{n}}} +\mathrm{1}\right\}^{{n}} =\left(\alpha+\beta−\gamma+\mathrm{1}\right)^{{n}} \\ $$$$\frac{{S}}{\left\{{S}^{\frac{\mathrm{1}}{{n}}} +\mathrm{1}\right\}^{{n}} }=\frac{\left(\alpha+\beta−\gamma\right)^{{n}} }{\left(\alpha+\beta−\gamma+\mathrm{1}\right)^{{n}} }=\left(\frac{\alpha+\beta−\gamma}{\alpha+\beta−\gamma+\mathrm{1}}\right)^{{n}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\alpha+\beta−\gamma+\mathrm{1}}\right)^{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}}{\left\{{S}^{\frac{\mathrm{1}}{{n}}} +\mathrm{1}\right\}^{{n}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\alpha+\beta−\gamma+\mathrm{1}}\right)^{{n}} \\ $$$$=\mathrm{0},\:{assume}\:\alpha+\beta−\gamma\geqslant\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{answer}\:\left({d}\right) \\ $$

Commented by mr W last updated on 12/Jan/20

is this the answer given in book?

$${is}\:{this}\:{the}\:{answer}\:{given}\:{in}\:{book}? \\ $$

Commented by TawaTawa last updated on 12/Jan/20

I really appreciate sir. God bless you sir.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by TawaTawa last updated on 12/Jan/20

This part answer is not in book sir. But your answer is in option.

$$\mathrm{This}\:\mathrm{part}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{in}\:\mathrm{book}\:\mathrm{sir}. \\ $$$$\mathrm{But}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{in}\:\mathrm{option}. \\ $$

Commented by mr W last updated on 12/Jan/20

and what do you think? actually it′s you who goes to the exam!

$${and}\:{what}\:{do}\:{you}\:{think}?\:{actually}\:{it}'{s} \\ $$$${you}\:{who}\:{goes}\:{to}\:{the}\:{exam}! \\ $$

Commented by TawaTawa last updated on 12/Jan/20

Your solution makes sense to me. So i will use the approach.

$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{to}\:\mathrm{me}. \\ $$$$\mathrm{So}\:\mathrm{i}\:\mathrm{will}\:\mathrm{use}\:\mathrm{the}\:\mathrm{approach}. \\ $$

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