The coefficient of t4 in the expansion of 1-t61-t 3 is -

( 1 t^61 t )^3=(1 t^6)^3·(1 t)^ 3 =(1 3t^6+3t^12 t^18)(1+3t+3·42!t^2+3·4·53!t^3+3·4·5·64!t^4+...+∞) ∴ Coefficient of t^4 =1×3·4·5·64! =3·4·5·64·3·2·1 =15 ...

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Byju's Answer

Standard XII

Mathematics

Binomial Theorem for Any Index

The coefficie...

Question

The coefficient of $t^{4}$ in the expansion of ${(\frac{1 - t^{6}}{1 - t})}^{3}$ is :

12

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10

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15

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14

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Solution

The correct option is C $15$
${(\frac{1 - t^{6}}{1 - t})}^{3} = (1 - t^{6})^{3} \cdot (1 - t)^{- 3}$
$= (1 - 3 t^{6} + 3 t^{12} - t^{18}) (1 + 3 t + \frac{3 \cdot 4}{2!} t^{2} + \frac{3 \cdot 4 \cdot 5}{3!} t^{3} + \frac{3 \cdot 4 \cdot 5 \cdot 6}{4!} t^{4} + . . . + \infty)$

$∴$ Coefficient of $t^{4}$ $= 1 \times \frac{3 \cdot 4 \cdot 5 \cdot 6}{4!}$

$= \frac{3 \cdot 4 \cdot 5 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}$

$= 15$

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The coefficient of t4 in the expansion of (1−t61−t)3 is :

The correct option is C 15(1−t61−t)3=(1−t6)3⋅(1−t)−3 =(1−3t6+3t12−t18)(1+3t+3⋅42!t2+3⋅4⋅53!t3+3⋅4⋅5⋅64!t4+...+∞) ∴ Coefficient of t4 =1×3⋅4⋅5⋅64! =3⋅4⋅5⋅64⋅3⋅2⋅1 =15

The coefficient of $t^{4}$ in the expansion of ${(\frac{1 - t^{6}}{1 - t})}^{3}$ is :