If the coefficients of the middle terms in the binomial expansions of and , , are equal, then the value of is: [2026]

Question

If the coefficients of the middle terms in the binomial expansions of ${(1 + α x)}^{26}$ and ${(1 - α x)}^{28}$ , $α \neq 0$ , are equal, then the value of is: [2026]

Smart Achievers · Accepted Answer

(4)

We have, ${}^{26}C_{13} α^{13} = {}^{28}C_{14} {(- α)}^{14}$

$\Rightarrow {}^{26}C_{13} α^{13} = \frac{28}{14} \times {}^{27}C_{13} α^{14}$

$\Rightarrow α = \frac{1}{2} \frac{{}^{26}C_{13}}{{}^{27}C_{13}} = \frac{1}{2} \frac{\frac{26!}{13! 13!}}{\frac{27!}{14! 13!}}$

$= \frac{1}{2} \times \frac{26!}{13!} \times \frac{14!}{27!}$

$= \frac{1}{2} \times \frac{26! \times 14 \times 13!}{13! \times 27 \times 26!} = \frac{7}{27}$

Solution

Q.
If the coefficients of the middle terms in the binomial expansions of ${(1 + α x)}^{26}$ and ${(1 - α x)}^{28}$ , $α \neq 0$ , are equal, then the value of is: [2026]

1 1

2 $\frac{14}{13}$

3 $\frac{27}{7}$

4 $\frac{7}{27}$

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Solution

Q. If the coefficients of the middle terms in the binomial expansions of (1+αx)26 and (1–αx)28, α≠0, are equal, then the value of is: [2026]

1 1

2 1413

3 277

4 727

Ans. (4) We have, C1326α13=C1428(–α)14 ⇒ C1326α13=2814×C1327α14 ⇒ α=12C1326C1327=1226!13!13!27!14!13! =12×26!13!×14!27! =12×26!×14×13!13!×27×26!=727

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Q.
If the coefficients of the middle terms in the binomial expansions of ${(1 + α x)}^{26}$ and ${(1 - α x)}^{28}$ , $α \neq 0$ , are equal, then the value of is: [2026]

2 $\frac{14}{13}$

3 $\frac{27}{7}$

4 $\frac{7}{27}$