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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

$t r y t o s o l v e b y y o u r s e l f . i t h i n k y o u$ $c a n s o l v e .$ $b u t t h e q u e s t i o n i s n o t a g o o d o n e . i t s a y s$ $o n l y α, β, γ > 0 . b u t α + β - γ c o u l d b e > 0,$ $= 0 o r < 0 w h i c h m a k e s t h e a n s w e r n o t$ $u n i q u e . s o a s s u m e α + β - γ ⩾ 0 .$
Commented by TawaTawa last updated on 12/Jan/20

$Sir, i asked questions i cannot solve .$ $I want to know the workings so that i can answer$ $correctly in exam .$ $You can help me if you are chanced sir .$ $Sorry to disturb you .$
Commented by mr W last updated on 12/Jan/20
$S=sum of all coef. of (ax+αy−γz)^n ⇒S=(α+β−γ)^n ⇒S^(1/n) =α+β−γ ⇒{S^(1/n) +1}^n =(α+β−γ+1)^n (S/({S^(1/n) +1}^n ))=(((α+β−γ)^n )/((α+β−γ+1)^n ))=(((α+β−γ)/(α+β−γ+1)))^n =(1−(1/(α+β−γ+1)))^n lim_(n→∞) (S/({S^(1/n) +1}^n ))=lim_(n→∞) (1−(1/(α+β−γ+1)))^n =0, assume α+β−γ≥0 ⇒answer (d)$
$S = s u m o f a l l c o e f . o f {(a x + α y - γ z)}^{n}$ $\Rightarrow S = {(α + β - γ)}^{n}$ $\Rightarrow S^{\frac{1}{n}} = α + β - γ$ $\Rightarrow {S^{\frac{1}{n}} + 1}^{n} = {(α + β - γ + 1)}^{n}$ $\frac{S}{{S^{\frac{1}{n}} + 1}^{n}} = \frac{{(α + β - γ)}^{n}}{{(α + β - γ + 1)}^{n}} = {(\frac{α + β - γ}{α + β - γ + 1})}^{n} = {(1 - \frac{1}{α + β - γ + 1})}^{n}$ $\lim_{n \to \infty} \frac{S}{{S^{\frac{1}{n}} + 1}^{n}} = \lim_{n \to \infty} {(1 - \frac{1}{α + β - γ + 1})}^{n}$ $= 0, a s s u m e α + β - γ ⩾ 0$ $\Rightarrow a n s w e r (d)$
Commented by mr W last updated on 12/Jan/20

$i s t h i s t h e a n s w e r g i v e n i n b o o k ?$
Commented by TawaTawa last updated on 12/Jan/20

$I really appreciate sir . God bless you sir .$
Commented by TawaTawa last updated on 12/Jan/20

$This part answer is not in book sir .$ $But your answer is in option .$
Commented by mr W last updated on 12/Jan/20

$a n d w h a t d o y o u t h i n k ? a c t u a l l y {i t}^{'} s$ $y o u w h o g o e s t o t h e e x a m!$
Commented by TawaTawa last updated on 12/Jan/20

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