binomial......die (1 Viewer)

[sab_JTL]

solve this for me .
practically i can't do maths any more

find the coeffient of x in the expansion: (2+9x)^4.(5+2/x)^5

where ^4 = to the power of 4
and . = times

show steps.

thank you

 

[spice girl]

Originally posted by sab_JTL
solve this for me .
practically i can't do maths any more

find the coeffient of x in the expansion: (2+9x)^4.(5+2/x)^5

where ^4 = to the power of 4
and . = times

show steps.

thank you

Hmm...
(2 + 9x)^4 = 2^4 + (4C1)*(2^3)*9*x + (4C2)*(2^2)*(9^2)*x^2 + (4C3)*(2)*(9^3)*x^3 + 9^4*x^4

(5 + 2/x)^5 = 5^5 + (5C1)*(5^4)*2*x^(-1) + (5C2)*(5^4)*(2^2)*x^(-2) + (5C3)*(5^2)*(2^3)*x^(-3) + (5C4)*(5)*(2^4)*x^(-4) + 2^5*x^(-5)

To get your coeff of x, we pair the coeffs of x with degrees like this:
1st set, 2nd set
1, 0
2, -1
3, -2
4, -3
Thus the coeff is:
(4C1)*(2^3)*9 * 5^5
+ (4C2)*(2^2)*(9^2) * (5C1)*(5^4)*2
+ (4C3)*(2)*(9^3) * (5C2)*(5^4)*(2^2)
+ 9^4 *(5C3)*(5^2)*(2^3)

= wotever that is...

why did i bother...?

 

[sab_JTL]

hmmm................
thanks.........
would something like this ever appear in hsc?

 

[Dumbarse]

why cant u do
nCr. a^n-r . b^n yeh that formula
and cause u want the coefficient of x ... make all the x terms into one term, (if possible), then put the power equal to 1. cause that is the power for the coeffiecient of just x, then once u find r, just solve it

will that work??

 

[spice girl]

Originally posted by Dumbarse
why cant u do
nCr. a^n-r . b^n yeh that formula
and cause u want the coefficient of x ... make all the x terms into one term, (if possible), then put the power equal to 1. cause that is the power for the coeffiecient of just x, then once u find r, just solve it

will that work??

what do you mean by "put the power equal to 1"?

the formula you stated is the binomial formula, it's the most basic formula and I have used it to expand each bracket.

this problem's a bit complicated because it involves 2 brackets, but all you have to do is to find all the coefficients and find all the combinations of pairing one term from one bracket from an appropriate term in another bracket. It's not that hard is it?

i dunno if it's goin to be asked. There's no easier way to do it (the number of terms unsimplified is 2^9)!

 

[Dumbarse]

ohh yeh
is this a binomial probability question??
well obviously i havent done this yet cause ive never seen it b4

 

[Dumbarse]

ohh ok dont worry i can do it, its just when u said 2^9!, i thought u went as factorial

your way just seemed complicated at first, i think my way would still work, its too hard to explain but

 

[Lazarus]

I think it would still work your way, Dumbarse. Might even be quicker. But you can't guarantee that the algebraic manipulation is going to be easier.

 

[spice girl]

it would definitely look simpler if i din have to use the ^ every time i wanted to raise by a power, and if i din need to do full working for that poor guy, i would just put this + this + ... + last term...

that might look simpler?

yeh dumbarse...show me your way...

 

[sab_JTL]

i think the second method would be faster, should work
any one care to give an example of how it actually works?/

 

[-=«MÄLÅÇhïtÊ»=-]

nah (Tr+1)/(Tr) is for greatest coef.

with this ques its trickier den the norm coz u have 2 expansions to do. But all u need to do is find the nth term, which in this case has power of 1 when (x)(y) multiply together.

So u find T(k+1) for the first expansion and and T(k+1) for 2nd expansion. Times them together, ie. add the powers of x and let them equal 1.

 

[spice girl]

Well, if no one realised, that's exactly what I did, but i wrote it all out first thing, because you'd eventually need to anyway...

And besides, it's easier to understand.

please, if there's a better method, illustrate it here instead of saying that you think it works...because i dun believe you..

 

[-=«MÄLÅÇhïtÊ»=-]

aiight fine, make me type!
spice gal, ino ur method works, but expanssion juz takes a lil bit longer...

aiight here i go!

To find the nth term, u use T(k+1)

(2+9x)^4

T(k+1)
= 4Ck(2)^(4-k).(9x)^k


(5+2/x)^5

T(k+1)
= 5Ck(5)^(5-k).(2/x)^k
= 5Ck(whatever)x^(k-1)

You're after the power of x, ie. 1
So equalling the powers:
1= k+(k-1)
So k=1

Sub k=1 into
{4Ck(2)^(4-k)}{5Ck(5)^(5-k)}
And u get the coef of x
= 100 000

 

[Lazarus]

Mmm, that's the good way - I used to hate expanding them.

 

[-=«MÄLÅÇhïtÊ»=-]

gotta save time in exam, no point writing out extra stuff dat wont get marks

 

[~ForAGoodCause~]

Originally posted by -=MLhtʻ=-




(5+2/x)^5

T(k+1)
= 5Ck(5)^(5-k).(2/x)^k
= 5Ck(whatever)x^(k-1)

how did ya get 5Ck(whatever)x^(k-1)

isnt it 5Ck(whatever)x^(-k) ?

plz explain im confused

 

[spice girl]

Originally posted by -=MLhtʻ=-
To find the nth term, u use T(k+1)

(2+9x)^4

T(k+1)
= 4Ck(2)^(4-k).(9x)^k


(5+2/x)^5

T(k+1)
= 5Ck(5)^(5-k).(2/x)^k
= 5Ck(whatever)x^(k-1)

You're after the power of x, ie. 1
So equalling the powers:
1= k+(k-1)
So k=1

Sub k=1 into
{4Ck(2)^(4-k)}{5Ck(5)^(5-k)}
And u get the coef of x
= 100 000

Firstly, there's no rationale in multiplying the (k+1)th term of the first expansion with the (k+1)th term of the 2nd. In fact, in this particular problem you don't.

Secondly (2/x)^k =/= (whatever)x^(k-1) whatever way you look at it.

Thirdly in your expression "{4Ck(2)^(4-k)}{5Ck(5)^(5-k)}"you forgot the (9x)^k from the first expansion, and the (2/x)^k in the second expansion.

You'd find the actual answer is quite a bit bigger than that.

There's no point finding a shorter way that doesn't work.

In fact the T(k+1) method only works in this case if you don't use k for the second expansion (use something like "l" or "a", not k!). Simplified, it ends in a sigma notation, where you scan thru all the solutions of k + l = 1.

Any objections?

 

[spice girl]

Ok, here's the T(k) method done my way. It's remarkably similar to the first method, but anyways:

(2+9x)^4.(5+2/x)^5

Let T(a,b) = (4Ca) * (2^(4-a)) * ((9x)^a) * (5Cb) * (5^(5-b)) * ((2/x)^b)

The expansion is:

Sigma{a = 0 to 4} ( Sigma{b = 0 to 5} ( T(a,b) ) )

Since T(a,b) = (some expression) * x^a * (x)^(-b)
Degree of x = 1
Thus a - b = 1
In the given ranges of a, b, we have several solutions:
a = 1, b = 0
a = 2, b = 1
a = 3, b = 2
a = 4, b = 3

Thus the coefficient is:
T(1,0) + T(2,1) + T(3,2) + T(4,3), when x=1
Converting this into numbers require writing it out, and that's wot I've done in my first post.

See? It's almost the same as my other soln!

 

[-=«MÄLÅÇhïtÊ»=-]

ouch, i guess they should post my previous post in the silly mistakes thread or make another thread called the silly idiotic mistakes.

Ok,

1.
"= 5Ck(5)^(5-k).(2/x)^k
= 5Ck(whatever)x^(k-1) "
that bit...i dunno, that was juz dumb of me.

2.
"Thirdly in your expression "{4Ck(2)^(4-k)}{5Ck(5)^(5-k)}"you forgot the (9x)^k from the first expansion, and the (2/x)^k in the second expansion."
Yes i should've included 9^k and 2^k



I'll try again, there's gotta be a better way den expanding it all out.
I made a mistake earlier by using k only, so now i'll use k and m.

(2+9x)^4

T(k+1)
= 4Ck(2)^(4-k).(9x)^k

(5+2/x)^5

T(m+1)
= 5Cm(5)^(5-m).(2/x)^m
=(whatever)x^-m

Addition of powers:
k-m=1

k can be from 0 to 4
m can be from 0 to 5
But since k-m=1,

k can only be from 1 to 4 and
m can only be from 0 to 3
So we have 4 possible combo's
(1,0), (2,1), (3,2), (4,3)
Spice gal, this is exactly wat u got up to, so im not tryn to steal ur thunder. But from this point on, ur first soln was obtained from expanssion. I'm gonna get it from my original method.


(2+9x)^4.(5+2/x)^5

f(k,m) = {4Ck(2)^4-k(9)^k}{5Cm(5)^5-m(2)^m}

f(1,0) = 900 000
f(2,1) = 12 150 000
f(3,2) = 29 160 000
f(4,3) = 13 122 000

Therefore coef. of x
= 1000(900+12150+29160+13133)
=55 332 000

Dat's a huge number, dunno if it's right.
Spice gal, can u plz do it ur way with the expanssion to check the answer?