Question

18 Term with x 5y18 in the expansion of (-3x +y) b. Constant term in the expansion of- z 15

Answer 1

(a) Here, we will use Binomial Expansion in order for finding term with $x^{5}y^{13}$ .

Binomial Expansion Formula is as : $(a + b)^{n}$ = $\binom{n}{r}(a)^{0} (b)^{r}$

So, let us get started for $(-3x + \frac{1}{4}y)^{18}$

$\implies$$\binom{18}{0}(-3x)^{18}(\frac{1}{4}y)^{0}$ +   $\binom{18}{1}(-3x)^{17}(\frac{1}{4}y)^1$ + $\binom{18}{2}(-3x)^{16}(\frac{1}{4}y)^2$ + $\binom{18}{3}(-3x)^{15}(\frac{1}{4}y)^3$ + ... + $\binom{18}{13}(-3x)^{5}(\frac{1}{4}y)^{13}$ + $\binom{18}{14}(-3x)^{4}(\frac{1}{4}y)^{14}$ + ... + $\binom{18}{18}(-3x)^{0}(\frac{1}{4}y)^{18}$   

Here, in above expansion we can see that 14th term having $x^{5}y^{13}$ ​​​​​​ term i.e. $\binom{18}{13}(-3x)^{5}(\frac{1}{4}y)^{13}$ =   $8568(-3x)^{5}(\frac{1}{4}y)^{13}$ ,

Thus , the 14th term having $x^{5}y^{13}$ term .

(b) Here , we are given   $(\frac{2}{z^{2}}- z)^{15}$ ,

we will expand this using Binomial expansion and

Let us consider the General Term for above Binomial is :

$T_{r+1}$ =  $\binom{15}{r}(\frac{2}{z^{2}})^{15 - r}(-z)^{r}$  

$T_{r+1}$ = $\binom{15}{r}(2)^{15-r}(-z)^{r}(z)^{2r-30}$   

$T_{r+1}$ = $\binom{15}{r}(2)^{15-r}(-z)^{3r-30}$   

Now , for a term in the expansion to be constant , the power of Z should be 0 .

So, $3r - 30 = 0 \enter \implies r = 10$ .

Therefore , the $11th$ term is the constant term in the expansion .

$T_{10+1} = T_{11}$

$\implies T_{11} = \binom{15}{10}(2)^{15-10} (-z)^{3*10 - 30}$

$\implies T_{11} = \binom{15}{10}(2)^{15-10} (z)^{0}(-1)$

$\implies T_{11} = -3003$