Question

..FOLLOWING EXPANSION,FIND THE TERM STATED.
(x-1/x)^6, constant term

Answer 1

@ganeshie8

Answer 2

@ikram002p

Answer 3

mmm what does that even mean :o

Answer 4

(x-1/x)^6, = (x-1) ^6 / x^6

use binomial theorm for (x-1)^6 :)

Answer 5

then expand :o
wait find (x-1)^6 from binomial theorm
(x-1 )^ 6 = x^6 + ......

so

(x-1) ^6 / x^6 = (x^6+ ...... -1) / x^6= 1 + stuff / x^6 - 1/x^6

Answer 6

so i guess its the first term :)

Answer 7

@Abmon98

Answer 8

i think it will be the 4th term

Answer 9

mmm @ganeshie8 i think the constant is 1 hehe
but the method might change its place :P

Answer 10

The given answer is -20

Answer 11

it is indeed -20 :
http://www.wolframalpha.com/input/?i=expand+%28x-1%2Fx%29%5E6

Answer 12

\[\large \left(x-\frac{1}{x}\right)^6\]

Answer 13

you want the power "3" on each terms so that they eat each other out :

\[\large \left(x\right)^{4-1}\left(\frac{1}{x}\right)^3\]

Answer 14

ok

Answer 15

that happens exactly when k = 3 :

\[\large \large \left(x-\frac{1}{x}\right)^6 = \sum \limits_{k=0}^6 \binom{6}{k} x^{6-k}\left(\frac{1}{x}\right)^k \]

Answer 16

-.-
why its not like this !
(x-1/x)^6 =(x^6-6x^5+15^4-20x^3+15x^2-6x+1) / x^6

Answer 17

\(k=3\) gives you 4th term in the expansion :

\[\large \binom{6}{3} x^{6-3}\left(-\frac{1}{x}\right)^3 \]

\[\large -\binom{6}{3} x^{3}\left(\frac{1}{x^3}\right)\]

\[\large -\binom{6}{3}\]

Answer 18

@ikram002p
(x-1/x)^6 = 1/x^6(x^2-1)^6

rught ?

Answer 19

wait my bad xD
i thought its( (x-1) / 6 )^6
xDDDDDD

Answer 20

="( lol

Answer 21

This gives me -20