modulo of series summation I have trouble with computing modulo.
First, I have a summation of series like this:
$$1+3^2+3^4+\cdots+3^n$$
And this is the formula which can be used to compute the series:
$$S=\frac{1-3^{n+2}}{1-3^2}=\frac{3^{n+2}-1}{8}$$
Then I want to compute $S \mod 1000000007$. But if $n$ is a large number, it's really hard for me to compute it. The only thing I know is $3^{n+2}-1$ divisible by 8 (n is an even number). Could anyone give me some good hint to solve this problem?

Update
My intention is computing $$M=\frac{3^{n+2}-1}{8} \mod 1000000007$$
For example: If $n=4000$ I must split $3^{4000+2}$ into $3^{40}3^{40}...3^{40}3^2$ and compute modulo for each part to improve speed like this:
$$(3^{40}\mod 1000000007)(3^{40}\mod 1000000007)...(3^{40}\mod 1000000007)3^2$$
But how can I compute M with the above idea.  

Update more
It seems related to inverse modulo. I think the problem was solved like this
$$I=\frac{(1000000007+1)}{8}=125000001$$
$$\frac{3^{n+2}-1}{8} \mod 1000000007=(3^{n+2}I-I)\mod 1000000007$$
Is it right?
The modulo value of a geometric sum 

1 + a + a^2 + ... +a^n mod m


can be calculated based on the identity (if the highest exponent is odd)

1 + a + a^2 + ... +a^(2x+1) = (1 + a)(1 + a^2 + (a^2)^2 + ... + (a^2)^x)


A complete algorithm which can deal with odd and even highest exponents looks like this in Mathematica: 

geometricSeriesMod[a_, n_, m_] := 
   Module[ {q = a, exp = n, factor = 1, sum = 0, temp},

   While[And[exp &gt; 0, q != 0],
     If[EvenQ[exp],
       temp = Mod[factor*PowerMod[q, exp, m], m];
       sum = Mod[sum + temp, m];
       exp--];
     factor = Mod[Mod[1 + q, m]*factor, m];
     q = Mod[q*q, m];
     exp = Floor[ exp /2];
   ];

   Return [Mod[sum + factor, m]]
]


Parameters:


a is the "ratio" of the series. It can be any integer (including zero and negative values).
n is the highest exponent of the series. Allowed are integers >= 0.
mis the integer modulus != 0


Note: The algorithm performs a Mod operation after every arithmetic operation. This is essential, if you transcribe this algorithm to a language with a limited word length for integers.
You can use the property of modulus,i.e., $$(a+b)\%m = (a\%m+b\%m)\%m$$

So to calculate 
$$S=1+3+3^2+3^3+...$$

you can use 

$$S\%M = (1\%M + 3\%M +3^2\%M +......)\%M$$

This will speed the process and even save you from integer overflow.