The coefficient of `t^(24)` in `(1+t^2)^(12)(1+t^(12))(1+t^(24))` is `^12 C_6+3` b. `^12 C_6+1` c. `^12 C_6` d. `^12 C_6+2`

Options

A. `""^(12)C_(6) + 3`
B. `""^(12)C_(6) + 1`
C. `""^(12)C_(6)`
D. `""^(12)C_(6) + 2`

Answer

c

Explanation

We have ,
`(1 - t)^(12) (t + t ^(12)) (1 -t^(24))`
` = {sum_(r=0)^(12) ""^(12) C_(r) (-t^(2))^(r)} {t - t^(25) + t^(12) - t^(36)}`
` = sum _(r=0)^(12) ""^(12)C_(r) (-t^(2))^(r){ t + t ^(12) - t ^(25) - t ^(36)}`
Coefficient of `t^(24)` = Coefficient of `t^(12)` in ` sum_(r=0)^(12) ""^(12)C_(r) (-t^(2)) ^(r)`
` = ""^(12)C_(6)` .