Probability that the sum of the numbers shown is a multiple of $5$ 
A regular die is rolled $n$ times such that $5|n$. Probability such that the sum of the numbers shown is a multiple of $5$ is given by $$\frac{a^n+b}{c\cdot d^n}.$$ Find $a,b,c$ and $d$.

What I thought is that we need to find coefficient of all those exponents of $x$ which are multiples of $5$ in the expansion of $(x+x^2+....+ x^6)^n$. Thereafter I am unable to solve it further.
 A: Using a little trickery with complex numbers, we can arrive at the sum of all coefficients of terms of degree a multiple of 5 of the above polynomial.
Consider the 5th roots of unity (i.e. the roots of $x^5=1$) and call them $(w_1,w_2,w_3,w_4,w_5)$
Observe:

*

*If k is a multiple of 5, then $w_1^k+w_2^k+w_3^k+w_4^k+w_5^k = 5$ (obvious)


*If k is not a multiple of 5, then $w_1^k+w_2^k+w_3^k+w_4^k+w_5^k = 0$ (not as immediately obvious, but not difficult to prove if you're familiar with De Moivre's Theorem or properties of Roots of Unity)
Let $P_n(x)=(x+x^2+....+ x^6)^n$ where n is a multiple of 5
Then we see that from our above observations:
Sum of coefficients of terms of degree a multiple of 5 $= \frac{P_n(w1)+P_n(w2)+P_n(w3)+P_n(w4)+P_n(w5)}{5}$
First, observe that when $w_i=1$ then $P_n(w_i)=6^n$
On the other hand, when $w_i$ is a nontrivial root of unity (i.e. not 1), then
$P_n(w_i)=(w_i+w_i^2+....+ w_i^6)^n$
$=(w_i(1+w_i+w_i^2+...+w_i^5))^n$
$=(w_i(1+w_i+...+w_i^4+1))^n$
$=(w_i(0+1))^n$
$=w_i^n$
$=1$ (since n is a multiple of 5)
Thus, $\frac{P_n(w1)+P_n(w2)+P_n(w3)+P_n(w4)+P_n(w5)}{5} =\frac{6^n+4}{5}$
Then the probability is:
$$\frac{6^n+4}{5*6^n}$$
