My 500 followers problem!

We have $P(2) = S = 500(2^{499}) + 495(2^{498}) + 490(2^{497}) + \ldots - 1990(2^1) - 1995$ : $\boxed{Eq.1}$

Multiplying 2 by Eq.1 in both the Left hand side, and well as Right Hand Side, we obtain:

=> $2S = 500(2^{500}) + 495(2^{499}) + 490(2^{498}) + \ldots - 1990(2^2) - 1995(2^1)$ : $\boxed{Eq.2}$

Subtract $\boxed{Eq.2}$ from $\boxed{Eq.1}$ to obtain:

=> $S = 500(2^{500}) - 5(2^{499}) - 5(2^{498}) - 5(2^{497}) - \ldots - 5(2^1) + 1995$

=> $S = 500(2^{500}) - (5)(2)[2^{498} + 2^{497} + 2^{496} + \ldots + 2^1 + 1] + 1995$

=> $S = 500(2^{500}) - (5)(2)[2^{499} - 1] + 1995$

=> $S = 500(2^{500}) - 5(2^{500}) + (5)(2) + 1995$

=> $S = 2005 + 495(2^{500})$

=> Therefore we have $a=2005; b=495; c=2; d=500; e=3002$ ; and $f=\boxed{2}$

This is simply an arithmetico-geometric progression , with

• First term of $AP = a_1 = -1995$ , Common difference $=d=5$ and 500th term $=a_n=500$

• First term of $GP = b_1 = 1$ , Common ratio $=r=x=2$ and 500th term $=2^{499}$

Sum of n terms of such progression is given by

$\dfrac{a_1b_1}{1-r}+\dfrac{db_1r(1-r^{n-1})}{(1-r)^2}-\dfrac{a_nb_nr}{1-r}$

Substituting the values , $= \dfrac{(-1995)(1)}{1-(2)} + \dfrac{(5)(1)(2)(1-2^{500-1})}{(1-2)^2}-\dfrac{(500)(2^{499})(2)}{1-2}$

$= \dfrac{-1995}{-1} + \dfrac{(5)(2)(1-2^{499})}{1}-\dfrac{(500)(2^{500})}{-1}$

$=1995 + 10 + 500(2^{500})-5(2^{500})$

$=2005+ 495(2^{500})$

$a+b+c+d=2005+495+2+500=3002 = e$

$3002 \equiv 2 \pmod{500}$

$\Rightarrow f =\huge\boxed{\color{#3D99F6}{2}}$ (smallest value)

@Azhaghu Roopesh M @Kunal Joshi