How do you use the pascals triangle to expand #(x + 2)^5#?

Question

1585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-30T16:14:48

#(x+2)^5 = x^5+10x^4+40x^3+80x^2+80x+32#

In general:

#(a+b)^5 = ((5),(0))a^5+((5),(1))a^4b+((5),(2))a^3b^2+((5),(3))a^2b^3+((5),(4))ab^4+((5),(5))b^5#

where #((5),(k)) = (5!)/((5-k)!k!)#

These binomial coefficients are found as a row of Pascal's triangle:

enter image source here

Write out the row of Pascal's triangle that begins #1, 5#...

#1, 5, 10, 10, 5, 1#

Write out powers of #2# up to #2^5#:

#1, 2, 4, 8, 16, 32#

Multiply the two sequences together:

#1, 10, 40, 80, 80, 32#

These are the coefficients we need:

#(x+2)^5 = x^5+10x^4+40x^3+80x^2+80x+32#