Question

Answer only if you know it.

Answer 1

Answer:

hope it may help you...................

Answer 2

Questions:

Find:

• (a) The coefficient of ${x}^{7}$ in the expansion of ${(a{x}^{2}+\frac{1}{bx})}^{11}$.

• The Coefficient of ${x}^{-7}$ in the expansion of ${(ax-\frac{1}{b{x}^{2}})}^{11}$

Also, find the relation between a and b , sk that the Coefficients are equal.

Answer:

$\large\bold\red{(a)^{11}C_{5}\:{a}^{6}{b}^{-5}}$

$\large\bold\red{(b)^{11}C_{6}\:{a}^{5}\:{b}^{-6}}$

$\large\bold\red{ab=1}$

Step-by-step explanation:

Let's head to the Question (a).

It's given that,

${( a {x}^{2} + \frac{1}{bx}) }^{11}$

We have to find the coefficient of ${x}^{7}$.

Let's write it's general term, i.e., (r+1)th term.

$T_{r+1}=^{15}C_{r} {(a {x}^{2}) }^{11 - r} {( \frac{1}{bx}) }^{r} \\ \\ = >T_{r+1}=^{11}C_{r} \: {a}^{11 - r} {b}^{ - r} {x}^{22 - 3r}$

Now,

We need power of x = 7.

Therefore,

$22 - 3r = 7 \\ \\ = > 3r = 22 - 7 \\ \\ = > 3r = 15 \\ \\ = > r = 5$

Therefore,

Coefficient of ${x}^{7}=\bold{^{11}C_{5}\:{a}^{6}{b}^{-5}}$

Now,

In Question (b).

We have given,

${(ax - \frac{1}{b {x}^{2} }) }^{11}$

It's general , (r+1)th term is,

$T_{r+1}=^{11}C_{r} \: {(ax)}^{11 - r} {( - \frac{1}{b {x}^{2} } )}^{r} \\ \\ = > T_{r+1}=^{11}C_{r} {( - 1)}^{r} \: {a}^{11 - r} {b}^{ - r} {x}^{11 - 3r}$

Now,

We need the power of x = -7

Therefor,

We get,

$= > 11 - 3r = - 7 \\ \\ = > 3r = 11 + 7 \\ \\ = > 3r = 18 \\ \\ = > r = 6$

Therefore,

We get,

Coefficient of ${x}^{-7}=\bold{^{11}C_{6}\:{a}^{5}\:{b}^{-6}}$

Now,

It's given that both the coefficients are equal,

Therefore,

We get,

$= > ^{11}C_{5} {a}^{6} {b}^{ - 5} = ^{11}C_{6} {a}^{5} {b}^{ - 6}$

But,

We know that,

• $^{11}C_{5}=^{11}C_{6}$

Therefore,

We get,

$= > {a}^{5} {b}^{ - 6} = {a}^{6} {b}^{ - 5} \\ \\ = > {a}^{6 - 5} = {b}^{ - 6 - ( - 5)} \\ \\ = > {a}^{1} = {b}^{ - 1} \\ \\ = > a = \frac{1}{b} \\ \\ = > ab = 1$