The coefficient of x3 in the expansion of 1−x+x25, is

The general term in the expansion of 1−x+x25 is5!r!s!t!(1)r(−x)sx2t,where r+s+t=5=5!r!s!t!(−1)sxs+2t       ...1For this term to contain x3, we must have  s+2t=3But r+s+t=5∴ r=2+t,s=3−2tNow, t=0⇒r=2,s=3t=1⇒r=3,s=1Thus/ there are two terms containing x3.Putting r=2, s=3, t=0 and r=3, s=1, t=1respectively in (i) , , we obtain that the terms are 5!2!3!0!(−1)3x3 and 5!3!1!1!(−1)x3Hence, Coefficient of x3=−5!2!3!−5!3!=−10−20=−30