The coefficient of the middle term in the bin | vedclass

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Question

(C) The middle term in the expansion of $(1 + \alpha x)^4$ is the $3^{rd}$ term,given by $T_3 = ^4C_2 (\alpha x)^2 = 6 \alpha^2 x^2$. The coefficient

Vedclass · Accepted Answer

$-3/10$

Vedclass · Answer

(C) The middle term in the expansion of $(1 + \alpha x)^4$ is the $3^{rd}$ term,given by $T_3 = ^4C_2 (\alpha x)^2 = 6 \alpha^2 x^2$. The coefficient is $6 \alpha^2$.The middle term in the expansion of $(1 - \alpha x)^6$ is the $4^{th}$ term,given by $T_4 = ^6C_3 (-\alpha x)^3 = -20 \alpha^3 x^3$. The coefficient is $-20 \alpha^3$.According to the problem,the coefficients are equal:$6 \alpha^2 = -20 \alpha^3$Since $\alpha 
eq 0$ (otherwise the middle terms would be constants,not involving $x$ in the same way),we can divide by $2 \alpha^2$:$3 = -10 \alpha$$\alpha = -3/10$.

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + \alpha x)^4$ and of $(1 - \alpha x)^6$ is the same if $\alpha$ equals

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