[/quote]
,
Prove that
be positive integers. Maximize 
if 
.
and 
be four prime numbers such that 
Prove that the sum of the four prime numbers is divisible by 
.
and 
and one starting number from the set 
used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number 
is reached or exceeded.
Show that Bob cannot win.
is given. Over the side 
draw an equilateral triangle 
on the outside. The midpoint of the segment 
is 
and the midpoint of the side 
is 
.
.
and 
be integers with 
.
such that 
(i.e., the square on 
has 
,
,
to the centers of the opposite squares are concurrent.
be an acute triangle. Let 
be the reflection of point 
over the line 
and 
be the midpoint of segment 
.
and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle 
is a point on the line 
such that 
,
.
be a regular hexagon with sidelength s. The points 
and 
are on the diagonals 
and 
,
.
,
be nonzero real numbers with 
Determine all possible values of 
of positive integers such that
,
be the angles of a triangle, and let 
,
.
,![\[s(s - a)(s - b)(s -c) \geq 0.\]](http://latex.artofproblemsolving.com/1/3/4/1344a0cbbd4e1aa4bcf6a1e777501c62daff32b3.png)
.
and 
lie on sides 
and 
,
and ![\[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \]](http://latex.artofproblemsolving.com/f/8/c/f8c35912c531eb2dae8f62b74666b7757d2b9ba3.png)
Prove that line 
.
Y by
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?
Induction proof wasn't quite as pleasing for me.
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?
Induction proof wasn't quite as pleasing for me.
This post has been edited 1 time. Last edited by MathBot101101, Apr 21, 2025, 6:45 AM
Reason: im dum
Reason: im dum
![\[ \sum\limits_{i=1}^n \dfrac{i}{(i + 1)!} = 1 - \dfrac{1}{(n+1)!}. \]](http://latex.artofproblemsolving.com/2/8/8/288b06247fd042559aaa1e0c0c101b5b2eba7821.png)
This is indeed a simple induction exercise. As for combinatorial arguments, maybe we can rearrange this into ![\[ \frac{1}{(n+1)!} + \sum\limits_{i=1}^{n} \frac{i}{(i+1)!} = 1,\]](http://latex.artofproblemsolving.com/f/1/a/f1a775f27856ac90581ef10e3fb4f6faaee58a71.png)
but then I'm not sure how to get a probability that is clearly equal to the LHS and is also equal to 
.
gives that
(This is induction at heart.)
