what is the answer to (x+y) to the 7th power?? no work needed... just an answer?

Answer 1

x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7

Answer 2

Use pascal's triangle.

(x+y)^0 = 1
(x+y)^1 = 1 1
(x+y)^2 = 1 2 1
(x+y)^3 = 1 3 3 1
(x+y)^4 = 1 4 6 4 1
(x+y)^5 = 1 5 10 10 5 1
(x+y)^6 = 1 6 15 20 15 6 1
(x+y)^7 = 1 7 21 35 35 21 7 1

Now use the "sum of the coefficients"
x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 +y^7.

The sum of the exponents is "7" per term. The numbers that were derived from the triangle are the coefficients to each term.

Answer 3

Given that you only want the answer, the prior answers will do. However, I would suggest you review the Binomial theorem and the Pascal theorem expansion, so you can answer these type of questions on demand, any time, any place.

Cofficients are: 1, 7, 21, 35, 35, 21, 7, 1

Good luck!

Answer 4

sum k from 0 to 7 of [ x^k * y^(7-k) * (7 choose k) ]