Question

how to prove the following?

Answer 1

@vishweshshrimali5 will like to do this..

Answer 2

wait..

I have a got an idea....
let me check if i am right....

Answer 3

@saadi


Sorry but it will take a lot of time.......

I would post the answer here.. whenerver I get it.........

Answer 4

I thought it is similar to the expansion of \((1+x)^n\). But, I would have to bring it to that form.

Answer 5

are u allowed to Using calculus (Maclaurin series ) for solving this problem?

Answer 6

@vishweshshrimali5
u got it already..........:)

Answer 7

It is also like

\(\large
(1+x)^n = 1 + nx + \frac{(n)(n-1)}{2! x^2}...\)

Answer 8

@mukushla

Sorry I couldn't understand what u want to say.....

Answer 9

I am writing the general formula for \((1+x)^n\)

Answer 10

@mukushla no we cannot use calculus. it is problem from binomial series section.

Answer 11

so let n=-1/2 to get ur answer

Answer 12

i mean expand the binomial series \((1+x)^n \) for \(n=-1/2\)

Answer 13

but that would give even negative terms.

Answer 14

finally we will let x=-1/2 and negative terms will change to positive

Answer 15

yes.............

@mukushla \(\huge WONDERFUL\)

Answer 16

idea is yours...:)

Answer 17

:)
But \(\huge YOU\) used it ........ \(\huge :)\)

Answer 18

\[(1+x)^{-1/2}=1+\sum_{n=1}^{\infty} \frac{(-1)^n 1 \times 3 \times ... \times (2n-1)}{2^n n!} x^n\]

Answer 19

@saadi

show that its true and then let x=-1/2 it will gives \(\sqrt{2}=1+2y \)

Answer 20

Well @mukushla
Handling the case to u......

I have to leave...

Answer 21

its almost done....:)

Answer 22

Ok continue ur work

Answer 23

@mukushla is simply the best

Answer 24

He is really the \(\huge BEST\)

He has helped me in many of my questions

@mukushla

Answer 25

Oh....guys...thank u so much......:)