# Challenge 74: Summing from the Year 2014 to N

Share
Embed
• Published on Jan 3, 2019
• Congratulations to reynolds45, Zain Majumder, Vampianist3, Nicola C, Rishav Gupta, Kwekinator117, and adandap for successfully solving the last week's math challenge question! reynolds45 was the first person to solve the question.
Your support is truly a huge encouragement.
Every subscriber and every like are wholeheartedly appreciated.
Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
For more Weekly Math Challenges:

• LetsSolveMathProblems  5 months ago +10

A few announcements:
* Please watch the first 40 seconds. Thank you so much for all of your encouragement throughout 2018! As stated in the video, I'm truly blessed to have such a supportive and amiable audience.
** Due to a busy personal schedule, the solution to the last week's challenge will be posted on Friday or Saturday. You have my sincere apologies.
*** The second semester (once I survive the chaos of the first few weeks) looks to be a much less stressful one for me, and I will endeavor to post as many non-Weekly-Math-Challenge videos as time permits. (A plethora of integration problems, Putnam, and AMC/AIME is on my bucket list.) =)

• magnifecent 5 months ago

Are you in university?

• QUANTUM city 5 months ago +1

Thank you to explain us awesome and interesting questions of mathamatics 😘

• Miyuki Umeki 5 months ago

• Kyu Ho Lee 5 months ago +1

How about actually substituting pi/4 on x? because the functions are periodic, might easily cancel out each other...
+ why did he say finding answers, when the only answer is 2019???

• Ethereal Huldra 5 months ago

I think you should let us viewers have access to the work you put up on your videos. You can probably put a link in the description. That way viewers don't need to go to specific parts of the video to find how you solved certain parts of a problem.

• mika2b 5 months ago

Thank you for this challenge !

• UbuntuLinux 5 months ago

This is my solution: www.upsieutoc.com/image/20190103-110209.GKE08M

• adandap 5 months ago +1

Another week of summer holidays and another "oops, it was math challenge day". :)
First, thank you so much for this channel. I've really enjoyed getting my eye back in for maths.
Second, happy 2019 - which is also the answer to this (not very hard) problem. First write the LHS as Sum [sec(3x) * (sin(ix) cos(3x) - cos(ix) sin(3x) )] = sec(3x) Sum[sin(i-3)x]. Then multiply both sides by cos(3x). Then on the RHS you can use sin(3x)cos(ix) = 1/2 * [sin(3+i)x + sin(3-i)x] and the sum telescopes to give just six terms: sin(2011x)+sin(2012x)+ ... + sin(2016x). The LHS will have the same six terms iff N=2019.

• Rishav Gupta 5 months ago

The answer is 2019 by multiplying cos3x both sides we get on lhs
Sin(ix)cos(3x)-sin(3x)cos(ix)=sin((i-3)x)
Whre i is from 2014 to N
Which implies sereis is
sin2011x+sin2012x....sin((N-3)x)
On rhs we get
Sin3x+sigma i from 1 to 2013(2sin(3x)cos(ix))
2sin(a-b/2)cos(a+b/2)
=sin a-sin b
On cancelling terms we get on right side
sin2011x+sin2012x......sin2016x
Which implies
N-3=2016
N=2019

• Daulian Doge 5 months ago +1

First I multiply cos3x to both sides which convert all tan3x into sin3x. Then at the LHS we see “sum(i=2014 -> N) of (cos3xsin(ix)-sin3xcos(ix))” which equal to “sum(i=2014->N) of (sin(ix-3x))” and in the RHS we see “sum(i=1->2013) of (2sin3xcos(ix))” equal to “sum(i=1->2013) of (sin(ix-3x)-sin(ix+3x))”. In RHS I separate “sum(i=1->2013) of (sin(ix+3x)-sin(ix-3x))” into “sum(i=1->2013) of (sin(ix+3x)) negative sum(i=1->2013) of (sin(ix-3x))” and add both sides with “sum(i=1->2013) of (sin(ix-3x))” which neutralize itself at RHS, I combine “sum(i=1->2013) of (sin(ix-3x))” with “sum(i=2014->N) of (sin(ix-3x))” and turn it into “sum(i=1->N) of (sin(ix-3x))”. Now I expand the RHS:

RHS = sin3x + sum(i=1->2013) of (sin(ix+3x))
= sin3x+sin4x+sin5x+...sin2016x

-> sum(i=1->N) of (sin(ix-3x)) = sin3x+sin4x+sin5x+...sin2016x
-> N=2019

*Sorry for bad english btw :)) *

• Shubham Kumar Yadav 5 months ago +1

Shift all the terms containing tan 3x to one side, rearrange the terms a bit so that the formulae for summation from sin(x) to sin(nx) and similarly for cos x to cos nx can be used, simplify a bit to convert LHS and RHS in terms of tan, and then as the result is true for all x, thus, by direct comparison, the answer is N=2019.
HAPPY NEW YEAR😉

• Aswini Banerjee 5 months ago

The answer is as I guessed 2019.
At first multiply both side by cos(3x) the equation becomes [∑cos(3x)sin(ix)-sin(3x)cos(ix) from 2014 to N] = sin(3x)+∑2sin(3x)cos(ix) from 1 to 2013
Now cos(3x)sin(ix)-sin(3x)cos(ix)=sin{x(i-3)} and 2sin(3x)cos(ix)=sin(3x+ix)-sin(ix-3x)
So [∑sin{x(i-3) from 2014 to N]=sin(3x)+∑sin(3x+ix)-sin(ix-3x) from 1 to 2013
Now ∑sin{x(i-3)} from 2014 to N+∑sin(ix-3x) from 1 to 2013=sin(3x)+∑sin(3x+ix) from 1 to 2013
Or ∑sin(ix-3x) from 1 to N=sin(3x)+∑sin(3x+ix) 1 to 2013
Now substituting (i-3) and (i+3) by k (2 ks are different) we get
[∑sin(kx) from 1 to N-3]-sin(x)-sin(2x)=sin(3x)+ ∑sin(kx) from 4 to 2016
Or [∑sin(kx) from 1 to N-3]=sin(3x)+sin(2x)+sin(x)+∑sin(kx) from 4 to 2016
Or [∑sin(kx) from 1 to N-3]=∑sin(kx) from 1 to 2016
Now comparing upper limits of both sides we get N-3=2013
so N=2019

• JHawk24 5 months ago +3

It's 2019 because it's the new year and it is the only answer that I would expect. It is also in red above the question.

• Hiren Bavaskar 5 months ago

First things first, convert tan3x as sin3x/cos3x and hence multiply both sides by cos3x..On LHS,it turns out to be summation from i=2014 to N of sin(i-3)x...on RHS convert 2sin3xcos(ix) into [sin(i+3)x - sin(i-3)x ] + sin3x.. Now as u write the terms of the summation , the only things left are sin2011x + sin2012x+...+sin2016x. Compare the RHS by writing the terms of summation of LHS, so we compare the last term i.e sin(N-3)x = sin 2016x , therefore N-3=2016 i.e N=2019 is the answer and you are done !

• Hiren Bavaskar 5 months ago

Correction* ,the RHS was sin3x + summation from i=1 to 2013 of 2sin3xcos(ix) which gets converted to sin3x+[sin(i+3)x - sin(i-3)x]

• Beshoy Nabil 5 months ago +5

Multiplying both sides by cos(3x) then simplifying using the law of sin of sum of two angles, we get:
sum(i=2014 to N) of sin(i-3)x = sin(3x) + sum(i=1 to 2013) of sin(i+3)x - sum(i=1 to 2013) of sin(i-3)x
Rephrasing the index and using the telescoping series in the RHS, we get:
sum(i=2011 to N-3) of sin(ix) = sin(3x) + sum(i=2011 to 2016) of sin(ix) - sum(i=-2 to 3) of sin(ix)
Using the odd property of sine function, the last series vanishes with the sin(3x) term, so we get:
sum(i=2011 to N-3) of sin(ix) = sum(i=2011 to 2016) of sin(ix)
Since the identity is valid for all real x, we are obliged to have N-3=2016 so N=2019
HINT: It would be a bit harder if not all real values of x but only for x=a/b *pi for certain conditions of a and b. It would have then another values for N.

• Beshoy Nabil 5 months ago

KAITO
That's a special case because x-x=0 for all x but in our case sin(2017x) + sin(2018x) + ... will not vanish except for certain values of x not all real possible variables of x as your series.

• Reygan Dionisio 5 months ago +2

2019 is it

• Lucio Antonio Rosi 5 months ago +3

2019

• Vampianist3 5 months ago +4