Find the first 3 terms of the expansion (3-5x)^14.how do one goes about this?

1 Answer
Feb 23, 2018

#4782969-111602610x+1209028275x^2#

Explanation:

We have the Binomial Theorem:

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

Here, #((n),(k))# is read as "n choose k" and is equal to #(n!)/(k!(n-k)!)#, where #!# is the factorial function.

Here, #x=3, a=-5x, n=14#. Inputting:

#(3-5x)^14=sum_(k=0)^14((14),(k))3^(14-k)(-5x)^k#

We'll do the first three terms:

When #k=0#:

#((14),(0))3^(14-0)(-5x)^0#

#=(14!)/(0!(14-0)!)*3^14#

#=1*4782969#

#=4782969#

When #k=1#:

#((14),(1))3^(14-1)(-5x)^1#

#=(14!)/(1!(14-1)!)*3^13*(-5x)^1#

#=14*1594323*-5x#

#=-111602610x#

When #k=2#:

#((14),(2))3^(14-2)(-5x)^2#

#=(14!)/(2!(14-2)!)*3^12(-5x)^2#

#=91*531441*25x^2#

#=1209028275x^2#

And we keep going on till we get to #k=14#.