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1

There is a well-known formula for the sum of the first $n$ squares, but I don't want spoil your investigation, so I will give you some hints. First, compute some more terms of the sequence. Three or four more should do. Multiply all the terms by six, and factor the results. Notice that all of them are multiple of $n$ and $n+1$.


0

Yes. $$\frac{n(n+1)(2n+1)}{6}$$ It's the sum of the first $n$ positive square numbers.


2

The Brahmagupta–Fibonacci identity says that every product of two sums of two squares is a sum of two squares in two different ways. So you have $$ 3^2+4^2=5^2\text{ and }20^2+21^2=29^2 $$ so consequently $5^2\times 29^2=145^2$ must be a sum of two squares in two different ways.


1

Suppose $a_1^2 + b_1^2 = c_1^2$ and $a_2^2 + b_2^2 = c_2^2$ are any two solutions. Multiply the first equation by $c_2^2$ and the second equation by $c_1^2$ to get $(a_1c_2)^2 + (b_1c_2)^2 = (a_2c_1)^2 + (b_2c_1)^2 = (c_1c_2)^2$. If you want only Pythagorean triples that are relatively prime, then consider $(d, e)$ where $d$ and $e$ are relatively prime and ...


0

If an odd number $c$ is the hypothenuse of two different primitive Pythagorean triplets, then $c$ is composite, and its prime factors are of the form $4k+1$. The smallest vakue for $c$ is $65=5\cdot 13$. Note that $65=8^2+1^2=7^2+4^2$. The proof of this fact that I know involves Gaussian integers.


4

Pattern recognition is a useful skill in a mathematician, but it is not actually mathematically rigorous. There is no strictly correct answer, but, if you encountered a sequence in the wild, on some problem, the continuation would be merely a conjecture. Consider: What is the maximum number of regions you get by drawing all the chords formed by $n$ points ...


0

For every $c$, $$n+c=\frac{\Gamma(n+c+1)}{\Gamma(n+c)},$$ hence the recursion you arrived at can be rewritten as $$a_{n+1}=\frac{-a_n}{4(n+r+1)(n+r+\frac12)}=a_n\frac{(-1)^n4^n\Gamma(n+r+1)\Gamma(n+r+\frac12)}{(-1)^{n+1}4^{n+1}\Gamma(n+r+2)\Gamma(n+r+\frac32)}a_n,$$ which immediately leads to $$ ...


0

Hint: take $z(x) = y(x^\alpha)$ with $\alpha>0$. Then try to write the differential equation on $z$ and try to guess a good $\alpha$.


0

The condition that a probability density must integrate to 1 gives you that its normally distributed.


0

$41$ is the only number in $A$ that is greater than $40$. And $5$ is the only one that is lesser than $6$. EDIT: And $29$ is the only one that nobody has yet mentioned...


0

Hint: think about numbers that appear only once in the sequence. Then think again.


0

Actually, any continuation sequence is viable: a finite number of terms do not uniquely define any sequence! This "problem" is one that many intelligence tests proffer. For my money, it is merely a demonstration of the test writer's inability to handle logic.


1

Why it is (obviously) $$S_n=504n^3-4158n^2+10647n-8269$$


1

Up to sign $$(1276-1):3=425,$$ $$(425-1):3=142,$$ and so on


1

If you prefer recursion: $a_n=a_{n-1}+n-1+3$ with $a_1=3$.


1

$$~~~~~~3 ~~~~~~ 6 ~~~~~~ 10 ~~~~~~ 15 ~~~~~~ 21 \\3 ~~~~~~~~ 4 ~~~~~~~~~ 5 ~~~~~~ 6 \\~1 ~~~~~~~~ 1 ~~~~~~~ 1$$ So $f(x) = ax^2 + bx + c$ Now $f(1)=3,f(2)=6,f(3)=10.$ From this we can solve $a,b,c$


2

$$3\quad 3+\color{red}3=6\quad 6+\color{red}4=10\quad 10+\color{red}5=15\quad 15+\color{red}6=21\quad $$


1

If you don't recognize the pattern (Triangular numbers), you can look at the differences. They are 3,4,5,6. Since the differences are an arithmetic sequence, you know the series is a quadratic, so can start with $n^2$. Then, you can call your expression $an^2+bn+c$ and solve it by using 3 of the values from your sequence.


1

$f(n)=\binom{n+2}{2}=\frac{(n+1)(n+2)}{2}$ If you have a sequence where the differences between terms is lineal then it is going to be a quadratic polynomial.



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