Question

2+1/2+1/2_ _ _ _ to infinite times.
What will be the value of the series?

Answer 1

what are the few other terms??

Answer 2

@ashishthomas7 , your 3rd term is incorrect. Check that :)

Answer 3

thats it 2+1/2+1/2+1/2+........to infinte terms

Answer 4

it would be infinity then no?

Answer 5

2 + (1/2 + 1/2 + 1/2 + ..................+1/2+...........) ??
that would be infinity

Answer 6

nope..its not the answer that i have in my book..

Answer 7

whats the answer given?

Answer 8

1+2^1/2

Answer 9

Well, Use

\[2 + \frac{1}{2} (1+1+1.............\infty)\]
\[2 + \frac{1}{2}( \infty)\]
\[2+ \infty\]
\[\infty\]

This should be the answer unless you have mistyped the question or I have made some mistake

Answer 10

its not infinity.

Answer 11

i meant infinte times

Answer 12

i have typed the question correctly

Answer 13

1+1+1+ .......(infinite times ) = infinte

Answer 14

it should have been like this
\[ 2(1+1/2-1/8+1/16-5/128+7/256 + .... ) \]

try to add few terms, obviously you will get the answer is invalid

Answer 15

@ashishthomas7 , can you take a screenshot of the question and attach it here :)

Answer 16

\[\sqrt{6+\sqrt{6+\sqrt{6+\infty}}}\]its a type of this question actually edited the stuff..

Answer 17

Yes

\[y =\sqrt{6+\sqrt{6+\sqrt{6+\infty}}}\]
\[y^2 = 6+ y\]

Answer 18

That's all :D

Answer 19

You get a quadratic equation. Now solve and get your y value

Answer 20

@ashishthomas7

Answer 21

well now can u do that in this series 2+\[2+1/2+1/2+.........\infty\]

Answer 22

How ??

Answer 23

i think we hav to do like this :\[2+1/2+1/2+.....\infty\] =\[2+1/X=X\]

Answer 24

@ashishthomas7 , Check again what you wrote .

Answer 25

i dont have the book :( but the question iscertainly like this
at a certain level u will get it \[2x+1=x ^{2}\]

Answer 26

do you mean a repeated fraction?

Answer 27

yep

Answer 28

According to you,
\[2+ \frac{1}{2+1/2 +1/2+1/2 +.......} = 2+1/2 +1/2 +...\]

How is this possible??

Answer 29

notlike that

Answer 30

(2+1/(2+1/(2+1/(2+....

Answer 31

That's what you wrote @ashishthomas7

@eigenschmeigen , his question is different

Answer 32

\[2+1/2+1/2+1/2+..........x terms\] now guys try

Answer 33

the question is not what yopu think
it should be
\[2+1\div(2+1\div(2+......))\]

Answer 34

Yes Exactly. That's why I am asking for some source / screenshot of the question

Answer 35

that means it turns out to be
\[x ^{2}=2x+1\]
=> x=\[1+\sqrt{2}\]

Answer 36

yup thats the answer.

Answer 37

@ashishthomas7 , please check your question before writing :)

Answer 38

I meant next time

Answer 39

@shivam_bhalla sorry :)

Answer 40

It's Ok :D

Answer 41

is it



?