Sum of a series

Okay, so looking at the function, we have the following:

$T(n+1)=T(n)(\frac{7+(-1)^{n-1}}{4})$

Which means that if n is odd, then we have T(n+1)=2T(n) and if n is even, T(n+1)=3/2T(n). We compute the first and last few terms:

$1,2,3,6,9,18,27,54,...3^{1006},2 \cdot 3^{1006}$

So if we want to find the units digit we pair the numbers as the following:

$1+2+3+6+9+18+...3^{1006}+2 \cdot 3^{1006}=(1+3+9+27+...+3^{1006})+(2+6+18+54+...+2 \cdot 3^{1006})=3(1+3+9+27+81+...+3^{2006})=3+9+27+81+243+...+3^{1007}$

So we want to find the units digit of this sum. Notice that the units digit of powers of 3 occurs in a pattern: 3,9,7,1,3,9,7,1... and so on. Notice 3+9+7+1=20 which means, if we group by 4s, each group of 4's units digit is 0, therefore not having an overall effect on the sum's units digit. To see what I mean, because I know I make no sense sometimes, see the following:

$3+9+27+81+243+...+3^{1007} \equiv (3+9+7+1)+(3+9+7+1)+...+(3+9+7+1)+3+9+7 \equiv 20+20+20+...+20+19 \equiv 19 \equiv \boxed{9} \mod (10)$

$\begin{aligned} \frac{T(n+1)}{T(n)} & = \frac{7+(-1)^{n-1}}{4} \\ \Rightarrow T(n+1) & = \frac{7+(-1)^{n-1}}{4} T(n) \\ \Rightarrow T(1) & = 1 \\ T(2) & = \frac{7+(-1)^0}{4}T(1) = 2(1) = 2 \\ T(3) & = \frac{7+(-1)^1}{4}T(2) = \frac{6}{4}(2) = 3 \\ T(4) & = 2(3) = 6 \\ T(5) & = \frac{3}{2}(6) = 9 \\ T(6) & = 18 \\ T(7) & = 27 \\ T(8) & = 54 \\ T(9) & = 81 \end{aligned}$

Let the sum of the 2014 terms be $S$ , then we have:

$\begin{aligned} S & = \sum_{n=1}^{2014} T(n) \\ & = \color{#3D99F6}{1 + 2} + \color{#D61F06}{3 + 6} + \color{#3D99F6}{9 + 18} + \color{#D61F06}{27 + 54} + \color{#3D99F6}{81 + 162} + ... + \color{#D61F06}{3^{1006} + 2 \left( 3^{1006} \right)} \\ & = \color{#3D99F6}{3(1)} + \color{#D61F06}{3(3)} + \color{#3D99F6}{3(9)} + \color{#D61F06}{3(27)} + \color{#3D99F6}{3(81)} + ... + \color{#D61F06}{3 \left( 3^{1006} \right)} \\ & = \color{#3D99F6}{3(1+9)} + \color{#D61F06}{3(3+27)} + \color{#3D99F6}{3(81+729)} + ... + \color{#D61F06}{3\left( 3^{1004}+3^{1006}\right)} + \color{#3D99F6}{3 \left( 3^{1005} \right)} \\ & = \color{#3D99F6}{3(10)} + \color{#D61F06}{3(40)} + \color{#3D99F6}{3(810)} + ... + \color{#3D99F6}{3(10)\left( 3^{1004}\right)} + \color{#D61F06}{3 \left( 3^{1005} \right)} \end{aligned}$

$\begin{aligned} \Rightarrow S & \equiv 3^{1006} \pmod{10} \quad \quad \small \color{#3D99F6}{\text{Since 3 and 10 are coprimes, by Euler's Totient Theorem: } 3^{\phi 10} \equiv 3^4 \equiv 1 \pmod{10} } \\ & \equiv 3^{4 \times 251+2} \pmod{10} \\ & \equiv \boxed{9} \pmod{10} \end{aligned}$