binomial theorem...

[Greatness]

In the example (2x+3)^4
after subbing the values into the equation thing, where does the (⁴₀) go when it is multiplied by (2x)^4?

[stonecold]

For smaller questions, you can just use pascals triangle. 

                         1
                       1   1
                     1   2   1
                   1   3   3   1
                 1   4   6   4   1
               1   5  10  10  5   1

[Greatness]

how would you use that in relation to these questoins?? lol this is new to mme

[dptjandra]

Method 1:
(4C0)*(2x)^4*(3)^0 + (4C1)*(2x)^3*(3)^1 + (4C2)*(2x)^2*(3)^2 + (4C3)*(2x)^1*(3)^3 + (4C4)*(2x)^0*(3)^4

Method 2:
Look at Pascal's triangle - (4C0) = 1 ; (4C1) = 4; (4C2) = 6; (4C3) = 4; (4C4) = 1
So it becomes:
1*(2x)^4*(3)^0 + 4*(2x)^3*(3)^1 + 6*(2x)^2*(3)^2 + 4*(2x)^1*(3)^3 + 1*(2x)^0*(3)^4

Similarly, if you had something to the power of 2, you would have:
(2C0) = 1; (2C1) = 2; (2C2) = 1

[stonecold]

Firsty expand.

So

(2x)⁴3⁰ +(2x)³3¹ + (2x)²3² + (2x)¹3³ + (2x)⁰3⁴

= 16x⁴ +24x³ + 36x² + 54x + 81

You have 5 terms, so go to the fifth line of pascals triangle, and place each of the coefficients in front of the terms.

Therefore,

1 x 16x⁴ + 4 x 24x³ + 6 x 36x² + 4 x 54x + 1 x 81

= 16x⁴ + 96x³ + 216x² + 216x + 81

Also, in Pascals Triangle, each number is just the sum of the two numbers above it.

For bigger questions you will need to use combinatorics though.

[Greatness]

Ohh i get it now i didnt know that the first row of the triangle = 0.
If a quesiton asks you to use the binomial theorem, can you still use the pascal triangle method?

EDIT: Yay thanks again guys i jsut did a few quetsions and got them right