Question-77902

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 = 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 .