Coefficient of t 24 in ( 1 + t 2 ) 12 ( 1 + t 12 ) ( 1 + t 24 ) is [2003]

Question

Coefficient of $t^{24}$ in ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$ is [2003]

Smart Achievers · Accepted Answer

(4)

${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$

$= (1 + t^{12} + t^{24} + t^{36}) {(1 + t^{2})}^{12}$

$∴$ $Coeff. of t^{24} = 1 \times Coeff. of t^{24} in {(1 + t^{2})}^{12} + 1 \times Coeff. of t^{12} in {(1 + t^{2})}^{12} + 1 \times constant term in {(1 + t^{2})}^{12}$

$= {}^{12}C_{12} + {}^{12}C_{6} + {}^{12}C_{0} = 1 + {}^{12}C_{6} + 1 = {}^{12}C_{6} + 2$

Solution

Q.
Coefficient of $t^{24}$ in ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$ is [2003]

1 ${}^{12}C_{6} + 3$

2 ${}^{12}C_{6} + 1$

3 ${}^{12}C_{6}$

4 ${}^{12}C_{6} + 2$

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Solution

Q. Coefficient of t24 in (1+t2)12(1+t12)(1+t24) is [2003]

1 C612+3

2 C612+1

3 C612

4 C612+2