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Solution


Q.

The sum of all rational terms in the expansion of (1+21/3+31/2)6 is equal to __________.          [2025]

Ans.

(612)

Given, (1+21/3+31/2)6

∴   The general term of multinomial expansion is 6!x!y!z!(1)x(21/3)y(31/2)z

For terms to be rational 3 divides y and 2 divides z.

∴   

x y z
6 0 0
4 0 2
2 0 4
0 0 6
3 3 0
1 3 2
0 6 0

∴   Required sum = 6!6!0!0!+6!4!0!2!×3+6!2!0!4!×(3)2+6!0!0!6!(3)3+6!3!3!0!(2)+6!1!3!2!(2)×(3)+6!0!6!0!(2)2

= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612