Any term in the expansion of (2x+x71+3x2)5 is given by
a!⋅b!⋅c!5!(2x)a(x−7)b(3x2)c
where a+b+c=5...(i)
=2a(3)c(a!⋅b!⋅c!5!)xax−7bx2c
=2a(3)c(a!⋅b!⋅c!5!)xa−7b+2c
Now for term to be independent of x, we know
a−7b+2c=0...(ii)
Solving (i)&(\mathrm{ii}), we get
b=8c+5
Also, a,b,c∈1,2,3,4,5
So, a=1,b=1,c=3
Therefore, the term independent of x will be
21(3)3⋅1!⋅1!⋅3!5!=1080