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(A) The general term in the expansion of $(a+2b+4ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{\alpha! \beta! \gamma!} a^{\alpha} (2b)

Vedclass · Accepted Answer

$315$

Vedclass · Answer

(A) The general term in the expansion of $(a+2b+4ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{\alpha! \beta! \gamma!} a^{\alpha} (2b)^{\beta} (4ab)^{\gamma} = \frac{10!}{\alpha! \beta! \gamma!} a^{\alpha+\gamma} b^{\beta+\gamma} 2^{\beta} 4^{\gamma}$.We are given the powers of $a$ and $b$ as $7$ and $8$ respectively,so:$\alpha + \beta + \gamma = 10$ $(1)$$\alpha + \gamma = 7$ $(2)$$\beta + \gamma = 8$ $(3)$Adding $(2)$ and $(3)$,we get $\alpha + \beta + 2\gamma = 15$. Subtracting $(1)$ from this,we get $\gamma = 5$.Substituting $\gamma = 5$ into $(2)$ and $(3)$,we get $\alpha = 2$ and $\beta = 3$.The coefficient is $\frac{10!}{2! 3! 5!} \cdot 2^{\beta} \cdot 4^{\gamma} = \frac{10!}{2! 3! 5!} \cdot 2^{3} \cdot (2^2)^{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{2 \cdot 6} \cdot 2^{3} \cdot 2^{10} = 2520 \cdot 2^{13} = 315 \cdot 8 \cdot 2^{13} = 315 \cdot 2^{16}$.Thus,$K = 315$.

If the coefficient of $a^{7} b^{8}$ in the expansion of $(a+2b+4ab)^{10}$ is $K \cdot 2^{16}$,then $K$ is equal to .... .

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