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In the expansion of (1+x) (1-x2 ) (1+ 3x+ 3x2+ 1x3 )5, x 0, the sum of the coefficients of x3 and x-13 is equal to .
118
(1+ 3x+ 3x2+ 1x3 )5= (1+ 1x )15= (1+x)15x15
(1+x) (1-x2 )=(1-x)(1+x)2 ⇒ expression = (1-x)(1+x)17x15
Coeff. of x3 = coeff. of x18 in (1-x)(1+x)17 =0-17C17=-1
Coeff. of x-13 = coeff. of x2 in (1-x)(1+x)17 =17C2-17C1=136-17=119
sum =-1+119=118
Answer: 118
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