Explain this method of expanding....

[umer007]

(n+2)^3

=n^3+3(n^2)(2)+3(n)(2^2)+2^3

=n^3+6n^2+12n+8

 

Could sum 1 tell me how this method cud b used for other exponents such as (n+2)^4 or (n+2)^5. So plz xplain to me this method of expanding, plz show the pattern if there is one.

[The Rebel]

Looking some stuff up on binomial expansion should reveal some ways it is used.

 

It basically considers expanding the brackets into its individual terms: i.e. ...

 

[math](a+b)^3 = (a+b)(a+b)(a+b) = a^3 + 3a^2b + 3ab^2 + b^3[/math]

 

To remember how its done, you realise that the number of terms will be a more than the power. I.e. in the example above, a power of 3 gives 4 terms.

 

The formula for each term when expanding [math](a+b)^n[/math] is [math]nCr * a^{n-r} * b^r[/math].

 

Where r is the term in the series (starting from 0). [math]nCr = \frac{n!}{r!*(n-r)!}[/math]

 

Applying this above to [math](a+b)^3[/math], to find the 2nd term its, [math]3C1 * a^{3-1} * b^1=3a^2b[/math]

 

Note the powers of a and b in each term always add up to n

[umer007]

wt dz the C stand for in that equation?

 

so for example if we take (n+2)^4,

 

we put it in the eqn:

 

4C0*n^4*2^0 = n^4

 

4C1*n^3*2^1 = 8n^3

 

4C2*n^2*2^2 = 16n^2

 

4C3*n^1*2^3 = 32n

 

4C4*n^0*2^4 = 64

 

= n^4+8n^3+16n^2+32n+64

 

Wouldnt you get this as an answer by following that formula. I know thats not correct so could u plz point out my mistakes. Duz the C mean to multiply the number infront of it by the number behind it.....or wtelse am i doing wrong.

 

Thx

[The Rebel]

You've almost got it except for the C bit. nCr is just a notation not a product, the C standards for combinations. You'll probably find nCr on a decent calculator next to nPr (permutations). If you do try entering 4 => nCr => 3 => ANS. You should get the answer 4.

 

To evaluate the co-efficient nCr you use the formula given above.

 

If you don't like using nCr, you can use Pascals triangle.

 

Have a look at this => http://www.davidscudder.com/pascal/binomial.html

[Primarygun]

The general term of the expansion :[math] nCr (a)^{n-r} (b)^r [/math]

[Primarygun]

oh The Rebel has mentioned it before.