Find the coefficient of $t^{20}$ in the expansion of $(t^{3} - 3 t^{2} + 7 t + 1)^{11}$

- 5654572342

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 6654572342

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 7654572342

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 8654572342

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 7654572342$
General term in the expansion of ${(t^{3} - 3 t^{2} + 7 t + 1)}^{11}$ is

$\frac{11!}{a_{1}! a_{2}! a_{3}! a_{4}!} {(t^{3})}^{a_{1}} {(- 3 t^{2})}^{a_{2}} {(7 t)}^{a_{3}} {(1)}^{a_{4}}$

$\Rightarrow a_{1} + a_{2} + a_{3} + a_{4} = 11 \Rightarrow a_{4} = 11 - a_{1} - a_{2} - a_{3}$ .....(i)

Also $3 a_{1} + 2 a_{2} + a_{3} = 20$

$\Rightarrow a_{3} = 20 - 3 a_{1} - 2 a_{2}$

Substituting the above value in equation (i), we get

$\Rightarrow a_{4} = 2 a_{1} + a_{2} - 9$

So, the coefficient of $t^{20}$ will be sum of

$\frac{11!}{a_{1}! a_{2}! (20 - 3 a_{1} - 2 a_{2})! (2 a_{1} + a_{2} - 9)!} {(- 3)}^{a_{2}} {(7)}^{a_{3}}$

such that $2 a_{2} + a_{1} \leq 20$ and $2 a_{1} + a_{2} \geq 9$

Thus coefficient is $- 7654572342$ .

Option C is correct.

Suggest Corrections

Similar questions

t^{4}

{(\frac{1 - t^{6}}{1 - t})}^{3}

t^{4}

{(\frac{1 - t^{6}}{1 - t})}^{3}

{(2^{x} + \frac{1}{4^{x}})}^{n}

\frac{T_{3}}{T - 2} = 7

Write the sum of the coefficients in the expansion of $(1 - 3 x + x^{2})^{111} .$

Express $x^{\frac{3}{4}} ({log}_{3} x)^{2} + {log}_{3} x - \frac{5}{4} = \sqrt{3}$ in terms of t where t = ${log}_{3}$ x.

Join BYJU'S Learning Program