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(D) For the binomial expansion $(1+\alpha x)^{26}$, the total number of terms is $27$ (which is odd), so the middle term is the $14^{th}$ term $(T_{14

Vedclass · Accepted Answer

$\frac{7}{27}$

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(D) For the binomial expansion $(1+\alpha x)^{26}$, the total number of terms is $27$ (which is odd), so the middle term is the $14^{th}$ term $(T_{14})$.$T_{14} = \binom{26}{13}(\alpha x)^{13}$, so the coefficient is $\binom{26}{13}\alpha^{13}$.For the binomial expansion $(1-\alpha x)^{28}$, the total number of terms is $29$ (which is odd), so the middle term is the $15^{th}$ term $(T_{15})$.$T_{15} = \binom{28}{14}(-\alpha x)^{14} = \binom{28}{14}\alpha^{14}x^{14}$, so the coefficient is $\binom{28}{14}\alpha^{14}$.Equating the two coefficients:$\binom{26}{13}\alpha^{13} = \binom{28}{14}\alpha^{14}$Since $\alpha 
eq 0$, we can divide both sides by $\alpha^{13}$:$\alpha = \frac{\binom{26}{13}}{\binom{28}{14}} = \frac{26!}{13!13!} \cdot \frac{14!14!}{28!}$$\alpha = \frac{26!}{28!} \cdot \frac{14!}{13!} \cdot \frac{14!}{13!} = \frac{1}{28 \cdot 27} \cdot 14 \cdot 14$$\alpha = \frac{196}{756} = \frac{7}{27}$.

If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$, where $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

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