,
that is defined as![\[
f(x)=\sum_{k=0}^{\infty}\bigg[ \frac{x+2^k}{2^{k+1}} \bigg]
\]](http://latex.artofproblemsolving.com/2/6/6/2663389b234754eb8eee666372d0b8d9b106e8a7.png)
Hence compute 
.
![\[
1+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\frac{1+2+3+4}{4!}+\dots
\]](http://latex.artofproblemsolving.com/9/4/8/9484cbccfd24b7119c9b9b3ef7f534e8f8b32e5e.png)
![\[
\int_{0}^{2022\pi}\sqrt{\frac{\cos{2x}-\cos{2\pi}(2022)}{\cos{2x}+1}}dx
\]](http://latex.artofproblemsolving.com/4/1/6/416b980d55c9cd9978061d451490bf70ca580c59.png)
that fulfills![\[
Q\left( \frac{y^2+2}{y}\right)-\frac{y^3+3y+1}{y}=\frac{y^3+8}{2y^2}
\]](http://latex.artofproblemsolving.com/0/d/6/0d621539963e157df7c2457e279315f4090c4af4.png)
Hence compute 
.
blue 
yellow 
green 
orange 
and purple 
Call an arrangement of blocks even if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement
is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is 
,
and 
.
.
to the equation 
be a given point inside quadrilateral 
.
and 
are located within ![\[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\]](http://latex.artofproblemsolving.com/d/b/3/db31d11ec24322724c0e0f1485aadf8864eabf7a.png)
Prove that 
if and only if 
.
be an integer. Find, with proof, all sequences 
of positive integers with the following three properties:
;
for all 
;
and 
(not necessarily distinct) for which 
,
.
![$[n] = \{1, 2, . . . , n\}$](http://latex.artofproblemsolving.com/4/f/f/4ff58e6e89a225ffa8db0ab50038fdadd9d4e2bc.png)
is a sequence 
such that each element of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
appears precisely one time as a term of the sequence. For example, 
is a permutation of ![$[5]$](http://latex.artofproblemsolving.com/c/4/2/c4216c69e041fca0904e8c5366d4480d1b2ca8b8.png)
.
be the number of permutations of 
is a perfect square for all 
.
.
has diameter 
of length 
.
lies tangent to 
and 
.
and 
,
is 
,
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. What is 
?
Y by
Penchick is playing a game! He starts by picking a positive integer 
and forms the sequence
![\[\dbinom{n}{1},\dbinom{n}{2},\dots,\dbinom{n}{n-2},\dbinom{n}{n-1}\]](//latex.artofproblemsolving.com/6/2/2/622773316bcfbce414950e2cf96017e951b72f95.png)
Let us denote the 
th term in this sequence as 
. Next, he picks some distinct positive integers 
. Then, he calculates 
, where 
denotes the amount of terms such that the highest power of 
that divides it is 
. His score is equivalent to the amount of distinct sums he can make given all possible . What is his score if he starts by picking 
?
and forms the sequence![\[\dbinom{n}{1},\dbinom{n}{2},\dots,\dbinom{n}{n-2},\dbinom{n}{n-1}\]](http://latex.artofproblemsolving.com/6/2/2/622773316bcfbce414950e2cf96017e951b72f95.png)
Let us denote the 
th term in this sequence as 
. Next, he picks some distinct positive integers 
. Then, he calculates 
, where 
denotes the amount of terms such that the highest power of 
that divides it is 
. His score is equivalent to the amount of distinct sums he can make given all possible . What is his score if he starts by picking 
?
is equal to 
,
.
,
s
for 
,
for all 
.
,
to be binary strings of length 
that can start with 
.
.
,
,
.
.
ways to arrange the 
or 
and assign 
.
,
.
ways to arrange the 1's, as for every remaining 
place values, you assign 
.
,
.
,
correspond to the numbers 
.
.
![\[ \binom{64}{k} = \frac{64}{k} \binom{63}{k-1}\]](http://latex.artofproblemsolving.com/8/f/e/8fed8d6dfcfba761576761b265fb3a992e6f1103.png)
By Lucas's theorem, the term 
is odd. Hence![\[ \nu_2 \left( \binom{64}{k} \right )=\nu_2 \left( \frac{64}{k} \right) = \nu_2(64) - \nu_2(k) = 6 - \nu_2(k) \]](http://latex.artofproblemsolving.com/9/c/0/9c0c66615f237c3900db06070133bab709e15493.png)
We now proceed by backwards induction on 
,
,
such 
.
for some 
.
,
,
,
meanwhile, then 
so there are 
in order to get a unique sum; adding any other terms is redundant to the 
term, which evaluates to 
.
,
