[/quote]
,
which is defined as all the natural numbers listed in order, side by side, behind a decimal point, without commas? If so, what is it?
and 
are primes, then can 
be a perfect square?
a prime number with 
and 
when 
.
else 
.
be real numbers such that 
.
be positive integers and 
and 
.
is a perfect square.
be a triangle with circumcentre 
.
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
.
be the circle passing through 
be a trapezoid such that 
lie on a circle with center 
,
is parallel to side 
.
and 
intersect at point 
,
.
.
,
and 
are positive integers and 
.
with 
such that ![\[ (a-1)(b-1)(c-1) \]](http://latex.artofproblemsolving.com/f/8/c/f8c33b59bdba1fdd12dec51f16a0b1b93e60754f.png)
is a divisor of 
is called good if there exists a permutation 
of the numbers 
,
and 
have different parities for every 
is a quadratic residue modulo 
for every 
is considered a quadratic residue modulo 
such that 
.
| 
with 
are positive integers and 
)
be a real numbers such that 
P
Where 
.
be non-negative real numbers. Prove that
Looking for an algebraic solution!
numebr on his list? Write its last two digits as your answer.
grid of square tiles. Robert chooses four of these squares uniformly at random. The probability that the centers of these four squares form a square themselves is 
where 
are coprime naturals. Write the largest 
digit odd factor of 
.
.
which touches one face of the cube and passes through all 
vertices of the face opposite to it. Find the integer closest to 
are distinct numbers in AP such that 
are also in AP. Find the smallest possible value of 
(in degrees).
that satisfy 
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.)
