calculate (x+a)^7?

Answer 1

Use Pascal's triangle.
the 7th row has coefficients 1, 7, 21, 35, 35, 32, 7, 1

so
(x+a)^7 = x^7 + (7 x^6 a) + (21 x^5 a^2) + (35 x^4 a^3) + (35 x^3 a^4) + (21 x^2 a^5) + (7 x^1 a^6) + a^7

Answer 2

The shortcut is to use the Binomial Theorem, which uses factorials and the trend that you start from exponent for x with 7 and decreasing incrementally by one till you get x^0 which is just 1.

Also, the exponent for the a term starts at a^0 = 1 and increases incrementally till you get a^7.

Then, evaluate the coefficients that are placed in front of each term based on the factorials from the Binomial Theorem. Please note the symmetry of the coefficients.

You should get ,

x^7a^0 + (7 x^6 a) + (21 x^5 a^2) + (35 x^4 a^3) + 35 x^3 a^4+ (21 x^2 a^5) + (7 x^1 a^6) + x^0a^7

The "long cut" is (x+a)(x+a)(x+a)(x+a)(x+a)(x+a)(x+a) which is one "FOILING" too many. Yikes!

Answer 3

woo... use binomial theorem to solve.

Answer 4

you can use binominal theorem
(x+a)^7=C(0,7)*x^7*a^0+C(1,7)*x^6*a^1+C(2,7)*x^5*a^2+
C(3,7)*x^4*a^3+C(4,7)*x^3*a^4+C(5,7)*x^2*a^5+C(6,7)*x^1*a^6+
C(7,7)*x^0*a^7=
=x^7+7x^6a+21x^5a^2+35x^4a^3+35x^3a^4+21x^2a^5+7xa^6+a^7

Answer 5

USE BINOMIAL EXPANSION