Question

please help

Answer 1

\[a _{n} = x^\frac{ 1 }{ 2^n }+ y^\frac{ 1 }{ 2^n }\] and b=\[x^\frac{ 1 }{ 2^n }- y^\frac{ 1 }{ 2^n }\] for all x belongs to N.

prove that \[a _{1}a _{2}a _{3}.........a _{n} = \frac{ x-y }{ bn }\]

Answer 2

we have to use concept of \[A.M \ge G.M \ge H.M\]

Answer 3

Sorry, but I have no clue how to do this type of problem. :/

Answer 4

i m sure nobody here does. lol

Answer 5

What kind of math is this? I may know someone that can help.

Answer 6

i have told what we have to use.

Answer 7

Oh, uhm maybe @iambatman, @Hero, or @Compassionate can help.

Answer 8

try this :

\[\large a _{1}a _{2}a _{3}\cdots a _{n} = \dfrac{1}{b^n} (a_1b)(a_2b)\cdots (a_nb) \]

Answer 9

what's this. And how u get that

Answer 10

find \(a_n\times b_n =... \)
then keep on simplifying using \((a-b)(a+b) = a^2-b^2\)

this method is equivalent to what ganeshie suggested,
but it doesn't use AM,GM,HM
thats why i stayed quiet for long....