Question

Find the 6th term of the expansion of (2p - 3q)11. a. -7,185,024p4q7 c. -7,185p4q7 b. -7,185,024p6q5 d. -7,185p6q5

Answer 1

$\bf (2p-3q)^{11}\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&&(2p)^{11}(-3q)^0\\ 2&+11&(2p)^{10}(-3q)^1\\ 3&+55&(2p)^9(-3q)^2\\ 4&+165&(2p)^8(-3q)^3\\ 5&+330&(2p)^7(-3q)^4\\ 6&+462&(2p)^6(-3q)^5 \end{array}$

the coefficient for the first term is 1, the next is 11 and so on... now, notice, the elements of the binomial, the 1st element starts off with 11, and every term it goes down by 1, the 2nd element starts off at 0, and goes up by 1 in each term.

now, to get the next coefficient, you simply, "get the product of the current coefficient and the exponent of the 1st element, and divide that by the exponent of the 2nd element in the next term".

for example, how did we get 165 for the 4th term.... well  (55*9)/3

how did we get 462 for the 6th term? well (330*7)/5.

and then you can just expand it from there.

Answer 2

Answer:  B.  $-7185024p^6q^5$

Step-by-step explanation:

The (r+1)th term in $(a+b)^n$ is given by :

$^nC_r(a)^{n-r}(b)^r$

The given binomial : $(2p - 3q)^{11}$

For the 6th term, we put r=6-1=5 , we get

$^{11}C_5(2p)^{11-5}(-3q)^5\\\\=\dfrac{11!}{5!(11-5)!}(2p)^6(-243q^5)\\\\=-dfrac{11\times10\times9\times8\times7\times6!}{6!5!}(64p^6)(243q^5)\\\\=-462\times(64p^6)(243q^5)\\\\=-7185024p^6q^5$

Hence, the 6th term of the expansion of $(2p - 3q)^{11}$ = $-7185024p^6q^5$