Question

Binomial Expansion : (1+x)^n type...

Answer 1

\[\frac{ 1 }{ 1-2x } +\frac{ 1 }{ 2+x } - \frac{ 2 }{ (2+x)^{-2} }\]

Answer 2

@cwrw238

Answer 3

what class u in, and what school u go to?

Answer 4

A level CIE exams for HSc.

Answer 5

The answer should be : 1+ 9/4 x + 15/4 x^2 ...

Answer 6

obtained : 1 + 5/2 x + -3x^2 :S

Answer 7

@Hero

Answer 8

\[( 1 -2x)^{-1} + (2+x)^{-1} - 2(2+x)^{-2}\]
is that correct ?

Answer 9

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

Answer 10

i need to find terms up to x^2

Answer 11

After Partial Fractions im asked to find binomial expansion up to x^2
the partial fractions results is correct but the binomial expansion result is not correct someone please help :D

Answer 12

Yeah, you should have clarified that at the beginning that way I could ping someone like @ganeshie8

Answer 13

\[(1-2x)^{-1} = 1 + (-1) (-2x) + \frac{ -1(-1-1) }{ 2! }x ^{2}\]

Answer 14

im not sure if this one is good how to make ( 1-2x)^-1 into ( 1+ x)^n form ?

Answer 15

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_33.pdf
Its Question number 9

Answer 16

@hartnn can you help me with that :D?

Answer 17

(....)^-2 <------what ?
you are done with partial fractions, right ?
denominators are 1-2x, x+2 and (x+2)^2

isn't it?

Answer 18

done with partial fractions and the denominators are 1-2x , x+2 and (2+x)^2

Answer 19

That's not what you posted originally @antoni7

Answer 20

right, so
\((1-2x)^{-1} = 1 + (-1) (-2x) + \frac{ -1(-1-1) }{ 2! }x ^{2}\)
is absolutely correct,
what further doubts u have ?

Answer 21

wait, i think there's an error

Answer 22

\[(1+\frac{ x }{ 2 }) ^{-1} = 1+ (-1) (\frac{ x }{ 2 }) + \frac{ -1(-1-1) }{ 2! }(\frac{ x }{ 2 })\]

Answer 23

\[(\frac{ x }{ 2 }) ^{2}\] i

Answer 24

last part is (x/2)^2 ..

Answer 25

(1+x)^n = 1+nx+n(n-1)x^2/2 (till x^2 term)

so, for (1-2x)^-1
n=-1, x=-2x

1+2x +4x^2
got this first ?

Answer 26

it should be 4x^2 and not just x^2 you got

Answer 27

yes got that

Answer 28

Here is what i got for all :
1 +2x +4x^2 +1 - x/2 +x^2/4 -1 + x - 3x^2/4

Answer 29

-2/(x+2)^2=-1/2 + x/2 -3x^2/8

Answer 30

did u miss the 2 ? in last 3 terms ??

Answer 31

-2/(x+2)^2= -1/2 * (1+x/2)^-2

Answer 32

last term :
\[- ( 1 + (-2) \frac{ x }{ 2 } + \frac{ -2 (-2-1) }{ 2! } (\frac{ x }{ 2 })^{2}\]

Answer 33

and u forgot the initial -1/2 .....

Answer 34

o.O

Answer 35

u got this :
"-2/(x+2)^2= -1/2 * (1+x/2)^-2"
right ?

Answer 36

\[-\frac{ 2 }{ (2+x)^{2} } = -2 (2+x)^{-2} = - (1+ \frac{ x }{ 2 })^{-2}\]

Answer 37

no^

Answer 38

is it -1/2 or -1 O.O ?

Answer 39

\(\large -2(2+x)^{-2} = -2 \times (2)^{-2} (\frac{2+x}{2})^{-2} = \frac{-1}{2} (1+\frac{x}{2})^{-2} \)

Answer 40

\((x+2)^2 = (2(1+x/2))^2=2^2 (1+x/2)^2= 4 (1+x/2)^2 \)

Answer 41

Omg!

Answer 42

got it now ?

Answer 43

wait

Answer 44

so i got : 3/2 + 2x + 31x^2/8 and the answerr should be as above

Answer 45

because 1st term is negative
-1/(1-2x) = -1-2x-4x^2
you forgot to consider the - *minus*

Answer 46

wait wait wait :D lets clear it all up :D

Answer 47

\[\frac{ 4 +5x -x ^{2} }{ (1-2x) (2+x)^{2} } = \frac{ 1 }{ 1-2x } + \frac{ 1 }{ 2+x } -\frac{ 2 }{ (2+x)^{2} }\]

Answer 48

\[\frac{ 4 +5x -x ^{2} }{ (1-2x) (2+x)^{2} } = \frac{ 1 }{ 1-2x } + \frac{ 1 }{ 2+x } -\frac{ 2 }{ (2+x)^{2} }\]

Answer 49

correct

Answer 50

so
\[(1-2x)^{-1} + ( 1 + \frac{ x }{ 2 })^{-1} - \frac{ 1 }{ 2 } ( 1 + \frac{ x }{ 2 })^{-2}\]

Answer 51

correct ?

Answer 52

1/(x+2) =....
i think here was the error

1/(x+2) = 1/2 -x/4+x^2/8

Answer 53

(1-2x)^-1 = 1+2x+4x^2
was correct

Answer 54

-2/(x+2)^2=-1/2 + x/2 -3x^2/8
was correct

Answer 55

now add

Answer 56

greaatttttttt yes 100% right answer for you :D

Answer 57

the problem was 1/2+x expansion.. jeez...

Answer 58

yeah ...big problem, silly mistakes can occur, but we finally nailed it :D

Answer 59

Super big thanks to all of you for correcting my silly mistakes.... :D