Question

Given that the constant term in the expansion $(a {x}^{3} - \frac{1}{x} ) {}^{12}$ is -1760. Find the value of a.

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expansion is

$\red{\rm :\longmapsto\: {\bigg[ {ax}^{3} - \dfrac{1}{x} \bigg]}^{12} \: }$

We know,

General Term in binomial expansion of (x + y)^n is

$\rm :\longmapsto\:\boxed{ \tt{ \: T_{r+1} = \: ^nC_r \: {x}^{n - r} {y}^{r} \: }}$

So, General Term of

${\rm :\longmapsto\: {\bigg[ {ax}^{3} - \dfrac{1}{x} \bigg]}^{12} \: \: is \: }$

$\rm :\longmapsto\:T_{r+1} = \: ^{12}C_r {[ {ax}^{3} ]}^{12 - r} {\bigg[ - \dfrac{1}{x} \bigg]}^{r}$

$\rm :\longmapsto\:T_{r+1} = \: ^{12}C_r {a}^{12 - r} {x}^{36 - 3r} \times \dfrac{ {(1)}^{r} }{ {x}^{r} }$

$\rm :\longmapsto\:T_{r+1} = \: {( - 1)}^{r} \: ^{12}C_r {a}^{12 - r} {x}^{36 - 3r - r}$

$\rm :\longmapsto\:T_{r+1} = \: {( - 1)}^{r} \: ^{12}C_r {a}^{12 - r} {x}^{36 - 4r}$

Since, it is given that, Constant term is - 1760.

So, Let we first find the Constant term, i.e. term independent of x.

So, to find the term independent of x, Substitute

$\red{\rm :\longmapsto\:36 - 4r = 0}$

$\red{\rm :\longmapsto\:- 4r = - 36}$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: r = 9 \: }}}$

So, term independent of x is

$\rm :\longmapsto\:T_{9+1} = \: {( - 1)}^{9} \: ^{12}C_9 {a}^{12 - 9} {x}^{0}$

$\rm :\longmapsto\:T_{10} = \: - \: ^{12}C_9 {a}^{3}$

So,

$\rm :\longmapsto\: \: - \: ^{12}C_9 {a}^{3} \: = \: - \: 1760$

$\rm :\longmapsto\: \: \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1} \times {a}^{3} \: = \: \: 1760$

$\rm :\longmapsto\: \: 220 \times {a}^{3} \: = \: \: 1760$

$\rm :\longmapsto\: \: {a}^{3} \: = \: \: 8$

$\rm :\longmapsto\: \: {a}^{3} \: = \: \: {2}^{3}$

$\rm \implies\:\boxed{ \tt{ \: a \: = \: 2 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_r \: = \: ^nC_{r - 1} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_0 \: = \: ^nC_{n} \: = \: 1}}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_1 \: = \: ^nC_{n - 1} \: = \: n}}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{^nC_r}{^nC_{r - 1}} = \frac{n - r + 1}{r} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_r \: + \: ^nC_{r - 1} \: = \: ^{n + 1}C_r \: }}$

Answer 2

Answer:

The value of a = 2

Step-by-step explanation:

Hope it helps you