# Ronelio’s Heavy Polynomial

The featured problem for today came from a young mathematician named Ronelio Barasi. He is an inc0ming 4th year High School of Dalandanan National High School. He came from a poor and broken family and grew up with separated parents. At this very young age, here are his few achievements in mathematics both local and international math challenges.

• 2014 MMC NCR Regional Individual – 3rd Place
• 2014 MMC NCR Sectoral Individual Category A – 1st Place 2014
• MMC NCR Sectoral Regional Team – 2nd Place
• 2013 MMC NCR Sectoral Regional Team – 1st place
• Bronze medalist, 2013 Environmental Math Cup International Level in Hong Kong
• Bronze medalist, 2013 King of Mathematics International Level in Hong Kong
• Silver medalist,  2013 Asia Cup Math Olympiad International Level Hong Kong

He is currently on training at the Mathematics Trainers Guiled Philippines in preparation for 2014 International Regions Math League to be held at Las Vegas, Nevada on May 28 to June 9 2014. He is planning to take Civil Engineering in U.P Diliman. Let’s take a look on a problem he shared.

Problem:

Let

$S=(x^3-x^2+x+3)^{2015}=a_0+(a_1)x+(a_2)x^2+(a_3)x^3+\dots+(a_{2014})x^{2014}+(a_{2015})x^{2015}$

Where, $\{a_0, a_1, a_2,\dots,a_{2015}\}$ are all integers.

Find the sum of  $a_0+a_2+a_4+\dots+a_{2012}+a_{2014}$

Solution:

Notice that, for every polynomial $P(x)=a_0+(a_1)x+\dots +(a_n)x^n$ of n degree.

We have $P(1)=a_0+a_1+a_2+\dots+a_n$

If n is odd, $P(-1)=a_0-a_1+a_2-\dots+a_{n-1}-a_n$

If n is even, $P(-1)=a_0-a_1+a_2-\dots-a_{n-1}+a_n$

Using this concept, we set $S=P(1)$,

So, $S =(x^3-x^2+x+3)^{2015}=(1-1+1+3)^{2015}$

$4^{2015}=a_0+a_1+\dots+a_n$………(1)

Now set $S=P(-1)$ note that 2015 is odd so,

$S=(x^3-x^2+x+3)^{2015}=(-1-1-1+3)^{2015}=0^{2015}$

$0=a_0-a_1+a_2-\dots+a_{n-1}-a_n$…….(2)

We add up (1) and (2) to obtain,

$4^{2015}+0=(a_0+a_1+a_2+\dots+a_n)+(a_0-a_1+a_2-\dots+a_{n-1}-a_n)$ $4^{2015}=2(a_0 + a_2 + a _4 + … + a_{n-1})$

Then, dividing both sides by 2,

$a_0+a_2+a_4+\dots+a_{2014}=2^{4029}$

Which is the required sum.