The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

The Proof is Trivial!

Scroll to see replies

Problem 509

Find ∑n=0∞n2n\displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}
Solution 509

Let Sn=∑n=0∞n2n=12+24+38+416+...∗ S_n = \displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n} = \frac{1}{2} + \frac{2}{4} + \frac {3}{8} + \frac {4}{16} +... *
Therefore 12Sn=14+28+316+432+...∗∗\frac{1}{2} S_n = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} +... **

Subtracting ** from *:

12Sn=(12+24+38+....)−(14+28+316+...) \frac{1}{2}S_n = (\frac{1}{2} + \frac{2}{4} + \frac{3}{8} +....) - (\frac{1}{4} + \frac{2}{8} + \frac{3}{16} +...)
  ⟹  12Sn=12+14+18+...=0.51−0.5=1  ⟹  Sn=2\implies \frac{1}{2} S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +... = \frac{0.5}{1-0.5} = 1 \implies S_n = 2
Original post by 16Characters....
Solution 509


Very nice! :u:

There's a neat trick to it at as well,

Spoiler

Original post by Zacken
Very nice! :u:

There's a neat trick to it at as well,

Spoiler



Nice. I'll finish my current graphs question and then try and think up some not involving matrices.
Original post by 16Characters....
Nice. I'll finish my current graphs question and then try and think up some not involving matrices.


I need to go practice some STEP, III 2000 is calling out to me. :rofl:
Original post by Renzhi10122
Problem 509

Let a,b,c a,b,c be the sides of a triangle. Prove that
ab+c+bc+a+ca+b+3abc(a+b)(b+c)(c+a)<2 \dfrac{a}{b+c}+\dfrac{b}{c+a}+ \dfrac{c}{a+b}+\dfrac{3abc}{(a+b)(b+c)(c+a)}<2


Spoiler

Original post by metaltron

Spoiler



Yep, quite nice I think.
Original post by Zacken
Problem 503

Given integers aa and bb satisfying ab<1\dfrac{a}{b} < 1, find integers cc and dd such that

ab+cd+acbd=1\displaystyle \frac{a}{b} + \frac{c}{d} + \frac{ac}{bd} = 1


Solution 503

Multiply out by bd.

ad+bc+ac=bd ad+bc+ac=bd
(b−a)(c+d)=2bc (b-a)(c+d)=2bc
We may assume that (a,b)=1 (a,b)=1 so c∣b−a c|b-a . Likewise, (c,d)=1 (c,d)=1 so b∣c+d b|c+d . 2 divides exactly one of b-a and c+d. Thus, we get two solution for c and d:
c=b−a,d=b+a c=b-a, d=b+a and
c=b−a2,d=b+a2 c=\dfrac{b-a}{2} , d=\dfrac{b+a}{2} and these do indeed work
Original post by Renzhi10122
Solution 503

Multiply out by bd.

ad+bc+ac=bd ad+bc+ac=bd
(b−a)(c+d)=2bc (b-a)(c+d)=2bc
We may assume that (a,b)=1 (a,b)=1 so c∣b−a c|b-a . Likewise, (c,d)=1 (c,d)=1 so b∣c+d b|c+d . 2 divides exactly one of b-a and c+d. Thus, we get two solution for c and d:
c=b−a,d=b+a c=b-a, d=b+a and
c=b−a2,d=b+a2 c=\dfrac{b-a}{2} , d=\dfrac{b+a}{2} and these do indeed work


I agree with you up to the before last line, love the solution although it was a little overkill!

If we have a=4,b=5  ⟹  ab=45a = 4, b=5 \implies \dfrac{a}{b} = \dfrac{4}{5}

So, c=b−a=1c = b-a = 1 and d=5+4=9d = 5+4 = 9 definitely works, I agree.

But c=5−42=12∉Zc = \dfrac{5-4}{2} = \dfrac{1}{2} \notin \mathbb{Z}
Original post by Zacken
I agree with you up to the before last line, love the solution although it was a little overkill!

If we have a=4,b=5  ⟹  ab=45a = 4, b=5 \implies \dfrac{a}{b} = \dfrac{4}{5}

So, c=b−a=1c = b-a = 1 and d=5+4=9d = 5+4 = 9 definitely works, I agree.

But c=5−42=12∉Zc = \dfrac{5-4}{2} = \dfrac{1}{2} \notin \mathbb{Z}


Very good point, ignore the last line :wink:
Problem 510*

Let MM be a point inside a triangle ABCABC. Prove that:

AB+BC+CA>MA+MB+MC+min(MA,MB,MC)AB+BC+CA>MA+MB+MC+min(MA,MB,MC)
(edited 8 years ago)
Original post by Renzhi10122
Problem 510*

Let MM be a point inside a triangle. Prove that:

AB+BC+CA>MA+MB+MC+min(MA,MB,MC)AB+BC+CA>MA+MB+MC+min(MA,MB,MC)


Presumaby A, B, C are the vertices of the triangle?
Original post by 16Characters....
Presumaby A, B, C are the vertices of the triangle?


Woops, yes they are, let me edit that...
Original post by Zacken
Problem 509

Find ∑n=0∞n2n\displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}

(∑n≥0qn)2=(∑n≥0qn)(∑n≥0qn)=∑n≥0∑0≤k≤nqkqn−k=∑n≥0∑0≤k≤nqn=∑n≥0(n+1)qn.\begin{aligned} \displaystyle \bigg(\sum_{n \ge 0}q^n\bigg)^2 & = \bigg(\sum_{n \ge 0}q^n\bigg) \bigg(\sum_{n \ge 0}q^n\bigg) = \sum_{n \ge 0} \sum_{ 0 \le k \le n} q^k q^{n-k} \\& = \sum_{n \ge 0} \sum_{ 0 \le k \le n} q^n = \sum_{n \ge 0}(n+1)q^n. \end{aligned}

Therefore (∑n≥0qn)2−∑n≥0qn=∑n≥0nqn\displaystyle \bigg(\sum_{n \ge 0}q^n\bigg)^2-\sum_{n \ge 0}q^n = \sum_{n \ge 0}nq^n

Set q=1/2q = 1/2 we have the required sum as 22−2=22^2-2 = 2.
Original post by Kummer
x


Nice! This is a pretty involved solution, you're pretty good at this?

Are you at uni? :smile:
Original post by Zacken
Nice! This is a pretty involved solution, you're pretty good at this?

Are you at uni? :smile:
Not that anyone would need this technique for this sum, but I think different solutions are good. Yes, I'm at uni.
(edited 8 years ago)
Original post by Zacken
Problem 509

Find ∑n=0∞n2n\displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}


Seems like a binomial expansion will work here:

4=(1−12)−2=1+22+34+48...=2Sn 4=(1-\frac{1}{2})^{-2}=1+\frac{2}{2}+\frac{3}{4}+ \frac{4}{8}...=2S_n
It follows that Sn=2 S_n=2
Original post by Renzhi10122
Seems like a binomial expansion will work here:

4=(1−12)−2=1+22+34+48...=2Sn 4=(1-\frac{1}{2})^{-2}=1+\frac{2}{2}+\frac{3}{4}+ \frac{4}{8}...=2S_n
It follows that Sn=2 S_n=2


I think you can just look at Sn and 0.5Sn, take one from the other, and be left with a geometric series... though I haven't put pen to paper with this one.

Posted from TSR Mobile
Original post by Krollo
I think you can just look at Sn and 0.5Sn, take one from the other, and be left with a geometric series... though I haven't put pen to paper with this one.

Posted from TSR Mobile


Yep, I saw, just thought I'd give my solution.
Original post by Renzhi10122
Yep, I saw, just thought I'd give my solution.


Good good. Always nice to have a good variety :ahee:

Posted from TSR Mobile

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you