Question

find the coefficient of X^6 using the binomial theorem given (1-3x)^2(x+2)^8

Answer 1

its fairly easy, just really long

Answer 2

and painful , boring algebra

Answer 3

(1 + 9x^2 - 6x) (x+2)^8
now find the coeff of x^6, and then add it to 9(coeff of x^4) -6(coeff of x^5)

Answer 4

coefficients to be found in (x+2)^8

Answer 5

u understand

Answer 6

ya thanx so i have to always expand one of the two given brackets? and i assume you expand the one with the smallest power?

Answer 7

not neccessarily lol

Answer 8

they could have high powers on both , thereby stoping you from expanding

Answer 9

they could also make the terms insides the brackets alot harder/tricker

Answer 10

it is convenient to expand one which is of a lower power like a square or cube,..and then progress

Answer 11

(4x-5)^10 (2x+10)^8

find coeffiecent of x^11 in that^

you dnt want to expand any of that :P

Answer 12

for an example

Answer 13

oh so eleceng... what do you recomend? than "him1618"

Answer 14

what do i do?

Answer 15

1,6,15,20, ...

Answer 16

now in elecengr's example u cant expan d...so uve got to do it the messy way

Answer 17

whew! how? please

Answer 18

its not too bad actually

Answer 19

just ask how can you get a power of x^11

Answer 20

multiply the coeff of x^0 in one to that of x^11 in the other, then add it to (coeff of x^1)(coeff of x^10) and keep adding like that..

Answer 21

well you can have a 10 and a 1 , or a 9 and a 2, or a 8 and a 3

Answer 22

etc etc

Answer 23

oh combinations...from one bracket suming with the other to give 11...simple right?

Answer 24

yes..bingo.and u keep on adding the products..got it?

Answer 25

ya thanx!!!! wow simple but tricky

Answer 26

but ther are also fractions....like [(2 / x)+(x ^2 / 2)^12]

Answer 27

[(2 / x)+(x ^2 / 2)]^12 sorry all to the power 12

Answer 28

The (n+1)th term in the expansion ofd thiswld be

(12Cn)*[(2/x)^(12-n) ]*[(x^2/2)^n]

wont it?

Answer 29

let me see

Answer 30

The expansion is
\[\sum_{0}^{n}[12Cn][2/x]^{12-n}[x^{2}/2]^{n}\]

Answer 31

i cant seem to get it.....i never thought Bi-theorem would b this hard when it comes to fractions i used to just change the variable x in the denominator so it becomes + and solve so if i was to find a constant behind x^6 i would use r as 7 (given my variable in the bracket has power 1,so it would then be -1 and subtract 1 from the 7 to get 6)

Answer 32

did u understand the expansion iwrote above?

Answer 33

no i dont get the C part

Answer 34

ok do u know what the general term of a binomial expansion of (a+b)^n is?

Answer 35

ya \[\sum_{r=0}^{n}((n! / r!(n-r)!).a^n-r.b^r\]

Answer 36

sorry a^n-r * b^r

Answer 37

yes....so nCr means the same as n!/r!(n-r)!

Answer 38

lol oh! wow i never knew....so ya your gen.eq above is true

Answer 39

so then in that expansion..please collect the powers of x and tell me the new form, where powers of x are all collected and powers of 2 are also collected

Answer 40

w8 a bit

Answer 41

i think im officially stuck,am i to expand from n=1 to n=12,i mean thats time consuming

Answer 42

no, you can use index laws :|

Answer 43

ok look up to my expansion above
now ill regroup it
\[\sum_{0}^{12}[12Cn][2]^{12-2n}[x]^{3n-12}\]

Answer 44

tyeh like that^

Answer 45

mmmm so should the n value i substitute be the same in the two variables or should i use any 2 varying numbers that will satisfy what i want

Answer 46

no, if you want the coeffiencent of say x^4 , then you set 3n-12 =4

Answer 47

yes...

Answer 48

nah thats not good

Answer 49

there actually is no coeffiecent of x^4 in this case , because when you solve that you dont get a natural number for n

Answer 50

you get a fraction

Answer 51

oh i see and then do i substitute the n value i got from 3n-12 =4 on both terms or do i have to solve 3n -12 to get its individual n value?

Answer 52

but say you wanted the constant term in the expansion (ie x^0)

Answer 53

then you would set 3n-12=0 , and solve to get n=4

then you take that value for n, and sub into the general coeffiecent of the expansion

Answer 54

so the constant term would be 12C4 2^(4)

Answer 55

oh il get the n value from 3n-12 and subst in the whole general eq...and again solve 12-2n and subst the n value....and the required variable and im done?

Answer 56

ull be given a condition..find the n-value from that and put it into the whole equation

Answer 57

o it sounds easy.....so i always ignore fractions...

Answer 58

no , but if you find n to be a fraction then it means that power (that was caused by that particular value of n ) doesnt exist , ie it is zero

Answer 59

its a bit hard to explain but you will find that n cant be fraction , because 12Cn is not defined unless n is a whole number and positive

Answer 60

try 12C(1/2) on your calculator, it will give an error, you cant choice (1/2) things out of 12 things lol

Answer 61

o so i keep that in mind.....what else is vital...i mean any tricks that one can fall in2

Answer 62

lol

Answer 63

just keep the equation of the general term in mind.....and the properties of the coefficients

Answer 64

thanx!!!