8/07/2014
Find
the term in the expansion (3x – 2y)5
that contains x2y3.
Scroll
down for the solution.
8/07/2014
Solution.
Since,
in general, the kth
term of (x + y)n
is nCk
– 1
x
(n
– k)·yk,
that for (3x – 2y)5
is
5Ck
– 1
(3x)
(5 – k)·(-2y)k
so
that with k
=
3 – because the exponent of y should be 3 – we have 5C3
– 1
(3x)
(5 – 3)·(-2y)3
= -720x2y3.