how to calculate 2^n modulo 1000000007 , n = 10^9

Question

what is the fastest method to calculate this, i saw some people using matrices and when i searched on the internet, they talked about eigen values and eigen vectors (no idea about this stuff)...there was a question which reduced to a recursive equation f(n) = (2*f(n-1)) + 2 , and f(1) = 1, n could be upto 10^9.... i already tried using DP, storing upto 1000000 values and using the common fast exponentiation method, it all timed out im generally weak in these modulo questions, which require computing large values

Answer 1

f(n) = (2*f(n-1)) + 2 with f(1)=1

is equivalent to

(f(n)+2) = 2 * (f(n-1)+2)
         = ...
         = 2^(n-1) * (f(1)+2) = 3 * 2^(n-1)

so that finally

f(n) = 3 * 2^(n-1) - 2

where you can then apply fast modular power methods.

Answer 2

Modular exponentiation by the square-and-multiply method:

function powerMod(b, e, m)
    x := 1
    while e > 0
        if e%2 == 1
            x, e := (x*b)%m, e-1
        else b, e := (b*b)%m, e//2
    return x

Answer 3

//Include library

const int mod = 1e9+7;

//Here base is assumed to be 2
int cal_pow(int x){
    int res;
    if (x == 0) res=1;
    else if (x == 1)    res=2;
    else {
        res = cal_pow(x/2);
        if (x % 2 == 0) 
            res = (res * res) % mod;
        else
            res = (((res*res) % mod) * 2) % mod;
    }
    return res;
}

int main(){
    int n, ans;
    cin >> n;
    ans = cal_pow(n);
    cout << ans;
    return 0;   
}