I need the answer and the procedure for (a+b)^n?

Question

Thank you so much if you can help me out with this. It really needs a lot. I know to use the Pascal's Triangle but i need the answer and a procedure on how to get the answer. Thank you.

Answer 1

Step 1:
First write the powers of 'a' in *descending* order:
a^n + a^(n-1)b + a^(n-2)b² + ... + a²b^(n-2) + ab^(n-1) + b^n

Step 2:
And write the powers of 'b' in *ascending* order:
1 + b + b² + ... + b^(n-2) + b^(n-1) + b^n

Step 3:
Multiply the first terms together, second terms, third terms, etc.
a^n + a^(n-1)b + a^(n-2)b² + ... + a²b^(n-2) + ab^(n-1) + b^n

Step 4:
Finally, take row 'n' of Pascal's triangle.
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
....

I'll just call these C(n, 0), C(n, 1), C(n, 2) ... C(n, n-2), C(n, n-1), C(n, n)

Put these in front of each of terms in step 3:
C(n, 0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b² + ... + C(n,n-2)a²b^(n-2) + C(n,n-1)ab^(n-1) + C(n,n)b^n

Okay, that's how to do it, but it is really confusing with all that notation. How about we do a real example with n = 5?

(a + b)^5

Step 1: Put 'a' in *descending* order (don't forget 1)
a^5 + a^4 + a^3 + a² + a + 1

Step 2: Put 'b' in *ascending* order (don't forget 1)
1 + b + b² + b^3 + b^4 + b^5

Step 3: Pair them up and multiply (you'll know you did this correctly if the exponents of each pair add up to 5 (n):
a^5 + a^4b + a^3b² + a² b^3 + ab^4 + b^5

Step 4: Use row 5 of Pascal's triangle: (Note: Some versions of the triangle might start with a row that is a single 1. This is known as row 0, so ignore it. If you aren't sure, look at the second number which will be the row number.)

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1 <--

Put each number in front (as the coefficient) of each term:

a^5 + 5a^4b + 10a^3b² + 10a² b^3 + 5ab^4 + b^5

Done!

Answer 2

(a+b)^2 = a^2 + 2a*b + b^2

Answer 3

You can use combination concept.
(a+b)^n = (a+b)(a+b)...(a+b)
If you expand it, then each term contains k 'a's and n-k 'b's.
So, all these combinations lead to
(a+b)^n = {i=0, n}∑C(n, i)(a^i)[b^(n-i)]