How do you expand #(r+s)^8#?

1 Answer
Dec 27, 2016

Answer:

#(r+s)^8 = r^8+8r^7s+28r^6s^2+56r^5s^3+70r^4s^4+56r^3s^5+28r^2s^6+8rs^7+s^8#

Explanation:

By the binomial theorem:

#(r+s)^8 = sum_(k=0)^8 ((8),(k)) r^(8-k) s^k#

where #((8),(k)) = (8!)/((8-k)! k!)#

Rather than calculate these binomial coefficients explicitly, we can read them from the #9#th row of Pascal's triangle (i.e. the row beginning #1, 8,...#)

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So:

#(r+s)^8 = r^8+8r^7s+28r^6s^2+56r^5s^3+70r^4s^4+56r^3s^5+28r^2s^6+8rs^7+s^8#