Binomial Theorem

Given the polynomial expression (1+x+x2+x3)100, we are interested in the coefficient of xr, denoted by ar. To find the sum of the products r⋅ar, we perform the following steps:
First, evaluate the polynomial at x=1 :
(1+1+12+13)100=(4)100
This shows that the sum of all coefficients,

300

∑

r=0

ar, equals 4100.
Next, differentiate (1+x+x2+x3)100 with respect to x :
‌

d

dx

((1+x+x2+x3)100)=100(1+x+x2+x3)99(1+2x+3x2)
Substitute x=1 into the derivative:

300

∑

r=0

r⋅ar‌=100(4)99(1+2⋅1+3⋅12)
‌=100(499)⋅6
Simplify the expression to find:

300

∑

r=0

r⋅ar=600(499)=150×4100
Since 4100=

300

∑

r=0

ar=S, we have:

300

∑

r=0

r⋅ar=150S
Thus,

300

∑

r=0

r⋅ar is equal to 150S.