Infinite Sum?

Before writing this post, I just want you to know that I'm still a student and my English isn't very good, so it might contain errors. :)--XmL
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According to my "more Interesting stuff" post. I mentioned that:
1/6+1/12+1/20
=1/(2*3)+1/(3*4)+1/(4*5)
=1/2-1/3+1/3-1/4+1/4-1/5
=1/2-1/5
=3/10
Now, let me ask you this: what's 1/6+1/12+1/20+1/30+1/42+............... to infinity?
well, let me start by breaking this down.
We know that 1/6=1/2*3=1/2-1/3, 1/12=1/3*4=1/3-1/4 and so on
so when I put this up: 1/2-1/3+1/3-1/4+1/4-1/5.......................
since -1/n+1/n=0
so 1/6+1/12+1/20+1/30+1/42+..............+ to infinity=1/2    (infinite sum!!!!)
this sum is in a form of 1/a*b+1/b*c+1/c*d+1/d*e+...............+ to infinity=1/a
is it true?
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Now let's explore the form (a/b)^n+(a/b)^(n+1)+(a/b)^(n+2)+..................+to infinity (where b>a)
Solve for S, S=(3/4)^2+(3/4)^3+(3/4)^4+(3/4)^5+................+to infinity         --------------(1)
multiply each side by 3/4:
3S/4=(3/4)^3+(3/4)^4+(3/4)^5+................+to infinity        -------------------(2)
(1)-(2):
S-3S/4=(3/4)^2
S/4=9/16
S=9/4

S-a*S/b=(a/b)^n
S*(b-a)/b=(a/b)^n
S=[b(a/b)^n]/(b-a)

For this reason,the solution of S=(a/b)^n+(a/b)^(n+1)+(a/b)^(n+2)+..................+to infinity (where b>a)
is S=[b(a/b)^n]/(b-a).
is it true?

New Discovery! (not really)

Since I have no life, I will share something that's pretty cool.

If y+mx=b where m>0, then the largest(maximum) value of xy (x times y) is b^2/4m.

I do have a proof of this but I don't know if it's useful or not. 

This theorem(?) really helps when you are finding the largest area of something. 

Proof-the Concurrency of medians of a triangle

Concurrency of Medians of a Triangle Theorem--The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.
 Proof:

Since CD bisects AB at point D and BE bisects AC at point F.
Point D and F are the midpoints
Connect D and F, giving us the midsegment of △ABC.
Using the Midsegment theorem
DF∥BC and DF=BC/2
DF/BC=1/2
Since DF∥BC and they're cut by transversal BF.
∠DFB≅∠FBC by the alternate interior angles theorem.
∠BOC≅∠DOF by the vertical angles theorem.
△DOF∼△COB By the AA Similarity Postulate.
DF/BC=FO/BO=DO/CO
Since BF/BC=1/2
FO/BO=1/2
Since BO+FO=BF
BF-BO=FO
(BF-BO)/BO=1/2
BF/BO-BO/BO=1/2
BF/BO-1=1/2
BF/BO=3/2
BO/BF=2/3 (Reciprocal Property)
BO=2BF/3 (Prove)
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More interesting stuff

1. 1/6+1/12+1/20=?

=1/(2*3)+1/(3*4)+1/(4*5)
=1/2-1/3+1/3-1/4+1/4-1/5
=1/2-1/5
=3/10

You might ask: why 1/2*3=1/2-1/3?
let's solve 1/2-1/3 first
1/2-1/3=1/6  
you might notice that the solution is the difference of the two denominators over the product of the two denominators.
so let a=2 b=3
1/a-1/b=(b-a)/ab
a+1=b
so  [(a+1)-a]/a(a+1)
=1/a(a+1)=1/a-1/(a+1)
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2. 1/(3*5)+1/(5*7)+1/(7*9)=?

(reminder: 1/2(x)=x/2)

=1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9)
=1/2(1/3-1/9)
=1/2(2/9)
=1/9

Why 1/3*5=1/2*(1/3-1/5)?
Using the same method
1/2(1/3-1/5)
=1/2(2/15)
=1/15
let a=3 a+2=5
1/2[(a+2)-a/a(a+2)]
=1/2[2/a(a+2)]
let k=2
1/k[k/a(a+k)]

1/3-1/5=2/15
1/a-1/(a+2)
=(a+2)/a(a+2)-a/(a+2)
=2/a(a+2)
let k=2
k/a(a+k)=1/a-1/(a+k)
so 1/k[k/a(a+k)]
=1/k[1/a-1/(a+k)]=1/a(a+k)

test:
1/5*7=1/5(5+2)
a=5 k=2
=1/2(1/5-1/7)

Geometry-P3

Problem↑

Solution (Several ways):

Make CE⊥AP, Connect BE.

Since m∠CEP=90° and m∠EPC=60

m∠ECP=30°

According to the 30-60-90 triangle theorem

2EP=PC

since 2BP=PC

EP=BP

m∠EBP=m∠BEP (EP's oppsite angle is congruent to BP's)

since m∠APC=60

m∠APB=120

m∠EBP=m∠BEP=(180-120)/2=30

Since m∠EBP=m∠ECP

BE=EC

m∠EBP=45-30=15

m∠BAP=60-45=15 (exterior angles theorem)

Since m∠EBP=m∠BAP

BE=AE=EC

Since AE=EC and ∠AEC=90

∠EAC=∠ACE=45

Since m∠ACE+m∠ECP=m∠C (angles addition postulate)

m∠C=45+30=75°

Geometry-P2

Problem:
An interior angle is missing in a convex polygon that has n sides, the sum of the other interior angles is 2570
find the value of n (how many sides this polygon has).

Solution:
The polygon interior angles theorem: 
The sum of the measures of the interior angles: (n-2)*180
Let x=the measure of the missing interior angle
(n-2)*180=2570+x
x=180n-2930
Since this is a convex polygon
0<x<180
therefore:
0<180n-2930<180
2930<180n<3110
16.2777<n<17.27777
Since n has to be a natural number
n has to be equal to 17
Therefore n=17

Interesting stuffs

1.   1/a+1/b=(a+b)/ab
Explanation:
1/a*b/b=b/ab
1/b*a/a=a/ab
a/ab+b/ab=(a+b)/ab

2.  
When x-y≥0
and y-x≥0
x=y
Explanation:
x≥y
y≥x
x=y (Only choice)          

Geometry-The Euclid Theorem Part1

The Euclid Theorem, also known as  射影定理, can be used when there's an triangle like this (Not really): 

                                                                                          

As can be seen, m∠ABC=90° and BD⊥AC

The following statements are true based on the given information.

1. (BD)^2=AD*DC，2.(AB)^2=AD*AC ，3.(BC)^2=CD*CA , 4.AB*BC=AC*BD

Proof for the the first statement: 

Since m∠BDC=90

m∠DBC+m∠C=90 (corollary to the triangle sum theorem)

and SInce m∠ABC=90

m∠A+m∠C=90   (corollary to the triangle sum theorem)

m∠DBC=m∠A

for the same reason

m∠A+m∠ABD=90

m∠A+m∠C=90

m∠C=m∠ABD

Since ∠C≌∠ABD and m∠BDA=m∠BDC=90

By the Angle-Angle Similarity Postulate

△ABD∽△BCD

Therefore AD/BD=BD/CD

(BD)^2=AD*DC

Geometry-P1

Since CF bisects∠ACB, BE bisect∠ABC
m∠CBO=m∠ABC/2
m∠BCO=m∠ACB/2
Using the Triangle Sum Theorem
m∠CBO+m∠BCO+m∠BOC=180
m∠ABC/2+m∠ACB/2+m∠BOC=180
(m∠ABC+m∠ACB)/2+m∠BOC=180
Since 180-m∠A=(m∠ABC+m∠ACB)
(m∠ABC+m∠ACB)=116
116/2+m∠BOC=180
58+m∠BOC=180
m∠BOC=122              

Inequality-P1

Problem:


Given that x>1
Find the lowest value of x+4/(x-1)


Solution:


x>1
x-1>0
let a=(x-1) b=4/(x-1)
Since (√a-√b)^2>0 and a≠b
a-2√ab+b>0
a+b>2√ab
By plugging in a and b
(x-1)+4/(x-1)>2√4(x-1)/(x-1)=4
(x-1)+4/(x-1)>4
Since (x-1)+4/(x-1)+1=x+4/(x-1)
x+4/(x-1)>5