we have
2n^3+3n^2 + n = n(2n^2+3n +1) = n(2n+1)(n+1)
= n((n-1) + (n+2))(n+1) = (n-1)n(n+1) + n(n+1)(n+2)
(n-1)n(n+1) is product of 3 consecutive integers and hence divisible by 6 and so is n(n+1)(n+2) and hence the sum is divisible by 6
some short and selected math problems of different levels in random order I try to keep the ans simple
we have
2n^3+3n^2 + n = n(2n^2+3n +1) = n(2n+1)(n+1)
= n((n-1) + (n+2))(n+1) = (n-1)n(n+1) + n(n+1)(n+2)
(n-1)n(n+1) is product of 3 consecutive integers and hence divisible by 6 and so is n(n+1)(n+2) and hence the sum is divisible by 6
As x^2 + y^ 2 = 1 we can chose x = \sin\, t , y = \cos\, t
3x + 4 y= 3 ]sin\, t + 4 \cos\, t
To convert 3x + 4 y= 3 ]sin\, t + 4 \cos\, t to the form A \sin (x+ t)
A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x
We can choose 3 = 5 \cos\, x and 4 = 5\ sin\, x (as 3^2 + 4^2 = 25 = 5^2)
= 5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)
It is maximum when \sin (x-t) = 1 and maximum value = 5
minimum when \sin (x-t) = -1 and minimum value = -5
We have n^2+ 3n + 2= (n+1)(n+2)
This is divisible by 2 for any n
For this to be divisible by 6 either n+1 or n+ 2 is divisible by 3 that is n should not be divisible by 3
The number of numbers that is divisible by 3 is \lfloor \frac{2024}{3} \rfloor = 674
So there are 664 numbers for which it is not divisible by 6 or there are 2024-674=-1350 numbers that are divisible by 6
We have
\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R
So \frac{2+3z-4z^2}{2 - 3z+ 4z^2} -1 \in R
Or \frac{6z}{2 - 3z+ 4z^2} \in R
Or \frac{2 - 3z+ 4z^2}{6z} \in R
Or \frac{2 - 3z+ 4z^2}{z} \in R
Or \frac{2}{z} -3 + 4z \in R
Or \frac{2}{z} + 4z \in R
Let z = x + iy and y\ne 0
So we have \frac{2}{x+iy} + 4(x+iy) \in R
Or \frac{2(x-iy)}{x^2+y^2} + 4(x+iy) \in R
Or \frac{-2y}{x^2+y^2} + 4y = 0 as imaginary part has to be zero
So as we have y is not zero dividing by y we get x^2+y^2 = \frac{1}{2} or |z|^2=\frac{1}{2}