If the coefficient of x30 in the expansion of (1+ 1x )6 (1+x2 )7 (1-x3 )8; x 0 is , then | | equals .
Answer
678
Solution
(1+ 1x )6= (1+x)6x6 = coefficient of x36 in
(1+x)6(1+x2)7(1-x3)8
General term: 6Cr1 7Cr2 8Cr3(-1)r3xr1+2r2+3r3, with r1+2r2+3r3=36
∵ r1+2r2 6+14=20 r3 6
r3=8 (r1+2r2=12): (r1, r2)=(0,6),(2,5),(4,4),(6,3) 7+315+525+35=882
r3=7 (r1+2r2=15): (1,7),(3,6),(5,5) -8(6+140+126)=-2176
r3=6 (r1+2r2=18): (4,7),(6,6) 28(15+7)=616
=882-2176+616=-678 | |=678
Answer: 678