Let P(x)=a0+a1x+a2x2+....+anxn be a polynomial in x where the coefficients a0,a1,a2,.....an are non-negative integers. If P(1)=25 and P(27)=1771769, then the value of a0+2a1+3a2+.....+(n+1)an is

The problem is asking for the value of the derivative of the polynomial P(x) at x=1.

The derivative of P(x) is P'(x) = a1 + 2a2x + 3a3x^2 + ... + nanx^(n-1).

We are asked to find P'(1) = a1 + 2a2 + 3a3 + ... + nan.

We know that P(1) = a0 + a1 + a2 + ... + an = 25 and P(27) = a0 + 27a1 + 729a2 + ... + 27^nan = 1771769.

Subtracting the first equation from the second gives:

26a1 + 728a2 + ... + (27^n - 1)an = 1771769 - 25 = 1771744.

Since all the coefficients are non-negative integers, the smallest possible value for the left side of the equation is 26a1.

Therefore, 26a1 <= 1771744, so a1 <= 68144.

Since a1 is a non-negative integer, the largest possible value for a1 is 68144.

Substituting a1 = 68144 into the first equation gives:

a0 + 68144 + a2 + ... + an = 25, so a0 + a2 + ... + an = 25 - 68144 = -68119.

Since all the coefficients are non-negative integers, the only way for this equation to hold is if a0 = a2 = ... = an = 0 and a1 = 68144.

Therefore, P'(1) = a1 + 2a2 + 3a3 + ... + nan = 68144 + 20 + 30 + ... + n*0 = 68144.