First slide

Binomial theorem for positive integral Index

Question

In the expansion of ${(7^{1 / 3} + 11^{1 / 9})}^{6561},$

Moderate

A

there are exactly 730 rational terms
B

there are exactly 5832 irrational terms
C

the term which involves greatest binomial coefficients is irrational
D

the term which involves greatest binomial coefficients is rational

Solution

General term is $^{6561} C_{r} 7 \frac{6561 - r}{3} 11^{\frac{r}{9}}$ .
To make the term free of radical sign, r should be a multiple of 9.
$∴ r = 0, 9, 18, 27, \dots, 6561 .$
Hence, there are 730 rational terms. The greatest binomial coefficients are
$^{6561} C_{\frac{6561 - 1}{2}} {and}^{6561} C_{\frac{6561 + 1}{2}} {or}^{6561} C_{3280} {and}^{6561} C_{3281}$ .
Now, 3280 are 3281 are not a multiple of 3; hence, both terms involving greatest binomial coefficients are irrational.

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