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Question

The coefficient of t32 in the expansion of (1+t2)12(1+t12)(1+t24) is

A
12C10+12C4
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B
12C5
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C
12C6
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D
12C7
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Solution

The correct option is A 12C10+12C4
(1+t2)12(1+t12)(1+t24)
=(1+t12)(1+t24)(1+t2)12
=(1+t12+t24+t36)expansion of (1+t2)12
=(1+t12+t24+t36)(1+12C1t2+12C2(t2)2+12C3(t2)3+12C4(t2)4+12C5(t2)5+12C6(t2)6+12C7(t2)7+12C8(t2)8+12C9(t2)9+12C10(t2)10+12C11+12C12(t2)12)
=(1+t12+t24+t36)(1+12C1t2+12C2t4+12C3t6+12C4t8+12C5t10+12C6t10+12C7t14+12C8t16+12C9t18+12C10t20+12C11t22+12C12t24)
Terms containing t32 is
=(t24)12C4t8+(t12)12C10t20
Co-efficients 12C10+12C4

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