Question:# Procedure on how to preform ( a+b)^n?

i need the answer and how to preform it. Teach me how to get the answer to this question. PLEASE!!

Step 1: First write the powers of 'a' in *descending* order: a^n + a^(n-1)b + a^(n-2)b? + ... + a?b^(n-2) + ab^(n-1) + b^n Step 2: And write the powers of 'b' in *ascending* order: 1 + b + b? + ... + 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)b? + ... + a?b^(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)b? + ... + C(n,n-2)a?b^(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 + a? + a + 1 Step 2: Put 'b' in *ascending* order (don't forget 1) 1 + b + b? + 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^3b? + a? 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^3b? + 10a? b^3 + 5ab^4 + b^5 Done!

use pascals triangle

a^n + nC1(a^(n-1))b + nC2(a^(n-2)b^2 + ... + nC(n-1)ab^(n-1) + b^n where nCr = n!/((n-r)!r!)

Just use the binomial theorem.

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