How do you use the binomial series to expand # (x+1)^20#?

1 Answer
Nov 9, 2016

Answer:

#(x+1)^20 = sum_(k=0)^20 ((20),(k)) x^(20-k)#

Explanation:

By the binomial theorem:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#

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

Putting #a=x#, #b=1# and #n=20# we get:

#(x+1)^20 = sum_(k=0)^20 ((20),(k)) x^(20-k)#

#color(white)((x+1)^20) = ((20),(0)) x^20 + ((20),(1)) x^19 + ... + ((20),(20))#

If you just want a single term from this expansion, then it is probably easiest to calculate directly.

For example, the fourth term is:

#((20),(3)) x^17 = (20!)/(17! 3!) x^17 = (20xx19xx18)/(3xx2xx1) x^17 = 1140 x^17#

If you want to write out every term, then it may be easier to write out Pascal's triangle as far as the #21#st row - hoping that you make no errors on the way. The #21#st row starts #1, 20, 190, 1140,...# consisting of the values:

#((20),(0)), ((20),(1)),...,((20),(20))#

conveniently providing all the coefficients you want.