#include <iostream>
#include <list>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <map>
#include <chrono>
#include <math.h>
using namespace std;

using lli = long long int;
using Vint = std::vector<int>;
using Vlli = std::vector<lli>;
using Wint = std::vector<Vint>;
using Wlli = std::vector<Vlli>;
using Vbool = std::vector<bool>;
using Wbool = std::vector<Vbool>;
using pii = std::pair<int, int>;
using pll = std::pair<lli, lli>;
template <class T>
using Vec = std::vector<T>;

constexpr int MOD = 1e9 + 7;
constexpr int INFi = 2e9 + 1;
constexpr lli INFl = (lli)(9e18) + 1;
const vector<pii> DXDY = {std::make_pair(1, 0), std::make_pair(-1, 0), std::make_pair(0, 1), std::make_pair(0, -1)};
constexpr char BR = '\n';

#define DEBUG(x) std::cerr << #x << " = " << x << '\n';
#define FOR(i, a, b) for(int (i) = (a); (i) < (b); ++(i))
#define FOReq(i, a, b) for(int (i) = (a); (i) <= (b); ++(i))
#define rFOR(i, a, b) for(int (i) = (b); (i) >= (a); --(i))
#define FORstep(i, a, b, step) for(int (i) = (a); i < (b); i += (step))
#define REP(i, n) FOR(i, 0, n)
#define rREP(i, n) rFOR(i, 0, (n-1))
#define vREP(ele, vec) for(auto &(ele) : (vec))
#define vREPcopy(ele, vec) for(auto (ele) : (vec))
#define SORT(A) std::sort((A).begin(), (A).end())
#define RSORT(A) std::sort((A).rbegin(), (A).rend())
// 座標圧縮 (for vector) : ソートしてから使うのが一般的 ; SORT(A) => COORDINATE_COMPRESSION(A)
#define COORDINATE_COMPRESSION(A) (A).erase(unique((A).begin(),(A).end()),(A).end())



template <class T> inline int argmin(std::vector<T> vec){return min_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline int argmax(std::vector<T> vec){return max_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline void chmax(T &a, T b){if(a < b) a = b;}
template <class T> inline void chmin(T &a, T b){if(a > b) a = b;}
template <class T> inline void reverseSORT(Vec<T> &Array){
  std::sort(Array.begin(), Array.end(), std::greater<T>());
}
inline int BitI(int k){return 1 << k;}
inline lli BitL(int k){return 1LL << k;}
inline void putsDouble(double d){printf("%.16lf\n", d);}
inline int toInt(const string &s){int res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline long long int toLong(const string &s){lli res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
template <class T> inline std::string toString(T n){
  if(n == 0) return "0";
  std::string res;
  if(n < 0){n = -n;while(n != 0){res += (char)(n % 10 + '0'); n /= 10;}
  std::reverse(res.begin(), res.end()); return '-' + res;}
  while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return res;
}

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



constexpr long double EPS = 1e-10;


int main(void){
    int n; long double p, q; scanf("%d%Lf%Lf", &n, &p, &q);
    /* P_i := i番目まで門があり, i 番目の門を通過する確率
       Q_i := i番目まで門があり, 元の場所に戻ってくる確率
       とすると
       k回, i番目に行って通過する確率は P_i * (pQ_i)^(k-1) * q だから (k ≥ 0)
       P_(i+1) = ∑_{k ≥ 0} P_i * (pQ_i)^(k-1) * q = pP_i / (1 - pQ_i)
       k回, i番目に行って戻る確率は pP_i * (pQ_i)^(k-1) * P_i だから (k ≥ 1)
       Q_(i + 1) = Q_i + p{P_i}^2∑_{k ≥ 0} (pQ_i)^k
                 = Q_i + p{P_i}^2 / 1 -pQ_i
    */
    Vec<long double> P(n + 1), Q(n + 1); P[0] = 1; Q[0] = 0;
    REP(i, n){
        long double oPQ = 1 - p * Q[i];
        if(abs(oPQ) <= EPS){
            P[i + 1] = 0;
            Q[i + 1] = 1;
        }
        else{
            P[i + 1] = q * P[i] / oPQ;
            Q[i + 1] = Q[i] + (p * P[i] * P[i]) / oPQ;
        }
    }
    printf("%.32Lf\n", Q[n]);

    return 0;
}