binomials (1 Viewer)

[Kawaii Mz Re]

could someone please help me find the constant term of (x^3 - 1/2x^2)^5
and please show working out

 

[clintmyster]

finding coefficient of x⁰

T_k+1 = 5Ck x (x³)^5-k)(-1/2x²)(^k)

taking out x terms

(x³)^5-k x (x^-2k) = x⁰

15 - 3k - 2k = 0
k = 3


Subbing k = 3 to find coefficient of x⁰, therefore omit x terms


T₃₊₁ = 5C3 x (1³)^5-3)(-1/2)(³)
T₄ = -5/4

 

[Trebla]

Note the sigma goes from r = 0 to r = 5...
(x³ - 1/2x²)⁵ = Σ ⁵C_r (x³)^{5 - r} (-1/2x²)^r
= Σ ⁵C_r (x)^{15 - 3r} (-1/2)^r (x)^-2r
= Σ ⁵C_r (x)^{15 - 5r} (-1/2)^r
We want the constant term, so we extract that term from the sum by letting 15 - 5r = 0 => r = 3
Hence the constant term is ⁵C₃ (x)⁰ (-1/2)³ = - 5/4

 

[addikaye03]

could someone please help me find the constant term of (x^3 - 1/2x^2)^5
and please show working out

(x^3 - 1/2x^2)^5

Tk+1=nCk a^n-kb^k
Tk+1=5Ck(x^3)^5-k(-1/2x^2)^k
Tk+1=5Ck(x^15-3k)(-1^k)(2x^-2k)

15-3k-2k=0
therefore k=3
T4=5C3(x^3)^2(-1/2x^2)^3
T4=-5C3/8
=-5/4

Pretty sure thats right, basically what u do is use the formula to find the indice of x, since it it constant the indice must =0 when u get it u re sub in eqn. Hope thats adequate explanation.

 

[Kawaii Mz Re]

the back of the text book says -1/1/4 but i dont know h ow to get it ><

 

[Kawaii Mz Re]

omgosh ty sooo much! =D lols thank again =D

 

[lyounamu]

could someone please help me find the constant term of (x^3 - 1/2x^2)^5
and please show working out

it's the fourth term.

so

5C3 (x^3)^2 (-1/2x^2)^3 = -5/4

 

[Timothy.Siu]

find the constant term of (x^3 - 1/2x^2)^5

first term has x^15, 2nd has x^10, 3rd has x^5 4th is constant...
4th term is:5C3x1^3 x -1/2^3=-5/4

 

[addikaye03]

finding coefficient of x⁰

5Ck x (x³)^5-k)(-1/2x²)(^k)

taking out x terms

(x³)^5-k x (x^2k) = x⁰

15 - 3k + 2k = 0
k = 15


Subbing k = 15

u went wrong cos when u raise 2x from the denominator it has a negative indice.

15-5k then = 0 etc etc

 

[addikaye03]

Jesus thats alot of posts for an answer and alot of differing methods.

 

[lyounamu]

Jesus thats alot of posts for an answer and alot of differing methods.

There are essentially same. It's just the length of each answer. I use table method which takes up about less than 1 minute.

 

[Kawaii Mz Re]

whats the '5C3' thingy? =S i ddont know y but our skool only taugh tthe pascal's triangle thingy... =[

 

[lyounamu]

whats the '5C3' thingy? =S i ddont know y but our skool only taugh tthe pascal's triangle thingy... =[

what do you mean?

5C3 means 5C3...????

 

[addikaye03]

There are essentially same. It's just the length of each answer. I use table method which takes up about less than 1 minute.

yea, well.. slightly differing. If i done it on paper it would take less than one minute using the method i used.

 

[Kawaii Mz Re]

what do you mean?

5C3 means 5C3...????

Where did the 'C' come from? =O

 

[lyounamu]

yea, well.. slightly differing. If i done it on paper it would take less than one minute using the method i used.

that's because the nature of the question given was not complicated.

Once you get into the question where the equation gets ridiculous, you will find that doing that way is little bit frustrating.

 

[addikaye03]

whats the '5C3' thingy? =S i ddont know y but our skool only taugh tthe pascal's triangle thingy... =[

its permutations and combinations/binomial theorem notation for 5x4x3x2x1/3x2x1 or 5x4... cant really explain it more than that

 

[addikaye03]

that's because the nature of the question given was not complicated.

Once you get into the question where the equation gets ridiculous, you will find that doing that way is little bit frustrating.

yer ino, i usually use your method, but i thought for a better explanation and more theoretical i thought i would show the formulation.

 

[Kawaii Mz Re]

its permutations and combinations/binomial theorem notation for 5x4x3x2x1/3x2x1 or 5x4... cant really explain it more than that

oooh! okay i get it now! ty soo much! thank you soo much everyone! =D hopefully ill get through this assessment thingy 2morrow =[

 

[Trebla]

My terrible version of Pascal's triangle lol:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

⁵C₃ corresponds to the 3rd co-efficient in the sixth row. It turns out that each term in Pascal's triangle can be expressed in combinatorial form (see your calculator for the combinatorial operation)...observe

⁰C₀
¹C₀ ¹C₁
²C₀ ²C₁ ²C₂
³C₀ ³C₁ ³C₂ ³C₃
⁴C₀ ⁴C₁ ⁴C₂ ⁴C₃ ⁴C₄
⁵C₀ ⁵C₁ ⁵C₂ ⁵C₃ ⁵C₄ ⁵C₅

Note that ⁿC_r = n! / [r!(n - r)!]