Question

Please help!
Find the 6th term of the expansion of (2p - 3q)^11.

a.-7,185,024p4q7
b.-7,185,024p6q5
c.-7,185p4q7
d.-7,185p6q5

Answer 1

You have to use Binomial Expansion Formula. If you have a binomial raised to a power say:\[(x + y)^{n} \] The expansion of this given by \[nC _{r} \times x ^{(n-r)} \times y ^{r}\]

Where n is the power and r is the no. of the term. In your problem n is 11. For 1st term r = 0 second term r=1 and so on.

For finding the 6th term r = 6-1 = 5 for your binomial the formula will be:

\[11C _{5} (2p)^{(11-5)} (-3q)^{5}\]
\[\frac{ 11\times10\times9\times8\times7 }{ 1\times2\times3\times4\times5} \times (2p)^{6} \times(-3q)^{5}\\[= -7185024 p ^{6}q ^{5}\]

Answer 2

\[\frac{ 11\times10\times9\times8\times7 }{ 1\times2\times3\times4\times5}\times(2p)^{6}\times(-3q)^{5}\] = \[-7185024 p ^{6}q ^{5}\]