MAA problem 1

There are two ways to solve this problem, but in both we must simplify the sum. Notice that this will be a geometric series, but since it is not infinite we must use a different equation.

${ S }_{ k }= a +ar+a{ r }^{ 2 }+\dots +a{ r }^{ n-1 }$

$r{ S }_{ k }= ar+a{ r }^{ 2 }+a{r}^{3}+\dots +a{ r }^{ n }$

$\left( 1-r \right) { S }_{ k }=a-a{ r }^{ n }$

${ S }_{ k }=\frac { a-a{ r }^{ n } }{ 1-r }$

The first way to solve the problem from here is to notice that $\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i = \cos { \frac { \pi }{ 3 } } +i\sin { \frac { \pi }{ 3 } }$ , which is generally well known to be equivalent to ${ e }^{ \frac { i\pi }{ 3 } }$ . Because of this, $a{ r }^{ 2015 }$ will be ${ e }^{ \frac { 2015 i\pi }{ 3 } }$ , which is equivalent to ${ e }^{ \frac { 2i\pi }{ 3 } }$ because the exponent is treated as an angle. Now that we are done with complicated exponent stuff, we put this back into our trigonometric expression, giving us $a{ r }^{ 2015 }=\frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i$ . It is easy to see that $a-a{ r }^{ 2015 }$ can now be represented as $\frac{1}{2}+\frac {\sqrt {3} }{2}i$ .

Now we divide this by $1-r$ , which is $\frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i$ in this case. To the solution all that is left is algebraic manipulation:

$\frac { \frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i }{ \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i }$

$\frac { \left( \frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i \right) \left( \frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i \right) }{ \left( \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right) \left( \frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i \right) }$

$\frac { -1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } i$

Which we then multiply by two to get $\boxed{-1+\sqrt{3}i}$

The other method is essentially the same. If we did not recognize the trigonometric trick earlier, writing out a few terms shows us that the terms repeat themselves on a period of four. Using this, we could find the appropriate value for $a{ r }^{ 2015 }$ and then finish the problem just as we did with the first method.

The other solutions here are great, but I wanted to submit a geometric "in your head" solution as well.

The multiplicative term in the recurrence relation is a primitive 6th root of unity, and it follows from arguments already submitted here that the terms of the sequence $a_n$ are simply all sixth roots of unity.

Now, the series $S_n$ is periodic because it's partial sums, thought of geometrically in the complex plane, form a hexagon with one vertex at $(0,0)$ (draw one iteration of this on graph paper to verify for yourself!). It follows that $S_{6k}=0$ for all $k \in \mathbb{N}$ , and since 2015 is 5 (mod 6), we have $S_{2015} = -\frac{1}{2}+\frac{\sqrt{3}}{2}$ .

The answer follows by multiplying this value by 2.

I did it by cube roots of unity!! In the summation ω^3=1 and ω^2+ω+1=0 helps alot..

1/2+sqrt(-3)/2 is omega(w).... Sq(w)+ w + 1 = O. a(n) = 1 S(2015) = 1 + w + sq(w)+ 1+w+sq(w)+......+1 + w=0(672 times) - sq(w) Sq(w) = 1/2-sqrt(-3)/2 Hence 2 *(-sq(w)) is the answer