Question

Find the eleventh term in the expansion: (3x – 2y) ^15

Answer 1

We will have the following:

We are given the expression:

$(3x-2y)^{15}$

Now, we know that the general binomial of the form:

$(a-b)^{15}$

Will be expanded as follows using Pascal's triangle:

$a^{15}-15a^{14}b+105a^{13}b^2-455a^{12}b^3+1365a^{11}b^4-3003a^{10}b^5+5005a^9b^6-6435a^8b^7+6435a^7b^8-5005a^6b^9+3003a^5b^{10}-1365a^4b^{11}+455a^3b^{12}-105a^2b^{13}+15ab^{14}-b^{15}$

Now, from this we can see that the 11th term is given by:

$3003a^5b^{10}$

Now, in our case a = 3x and b = 2y, thus:

$\begin{gathered} 3003(3x)^5(2y)^{10}=3003(243x^5)(1024y^{10}) \\ \\ =747242496a^5b^{10} \end{gathered}$

So, the 11th term is:

$747242496a^5b^{10}$