A problem in honour of solving 1000 problems

You can notice that every pentagonal number includes the privious one but you add a certan number So I wrote it as a sequence (1, 4, 7, 10...) [notice that 5=4+1 , 12=7+5] So we can concloude that any pentagonal number equals the sum off all n terms before it Using the rule sn=n/2 (2a+d (n-1)) , 1001=n/2(2+3 (n-1)), 3n^2-n-2002=0, n= 26 or n=-77/3, n=26

The simplest way to test whether a positive integer $x$ is a pentagonal number is :

$\Large n=\dfrac{\sqrt{24x+1}+1}{6}$

If $n$ is a natural number , then $x$ is the $n^{th}$ pentagonal number.

Here $x=1001$ , we get :

$\Large {n=\dfrac{\sqrt{24(1001)+1}+1}{6} \\ \Rightarrow n=\dfrac{\sqrt{24024+1}+1}{6} \\ \Rightarrow n=\dfrac{\sqrt{24025}+1}{6} \\ \Rightarrow n=\dfrac{155+1}{6} \\ \Rightarrow n=\dfrac{156}{6}} \\ \Rightarrow \huge\boxed{n=26}$

A method my 8th grade mathcounts teacher taught me is that pentagonal, triangonal, etc. numbers can be approximated with a combination of a square and triangles.

A triangular number is $n(n-1)/2$ A square is $n^2$

Thus a pentagonal number would be $n^2 + n(n-1)/2$ . Much easier than memorizing formulas, right?!

And n=26.

The Expression for the nth pentagonal number is n(3n-1)/2,

Equate n(3n-1)/2 = 1001 and this gives us the quadratic equation:

3n^2 - n - 2002 = 0

Solver for n

3n^2 - 78n + 77n - 2002 = 0

3n(n-26) + 77(n-26) = 0

(3n + 77)(n -26) = 0

The only solution permissible is n = 26

This is a pretty easy way to do it too, provided that the number is not enormous.

Incidentally the formula used in this script and that test are both provided in wikipedia.

Note that $1 = 1, 5 = 1 + 4, 12 = 1 + 4 + 7$ etc. So these numbers are in the form of $3n - 2$ .

A sum of an arithmetic progression is the (average of the first and last number) times (number of terms)

(example $1+2+3...+9 = (1+9)/2 \times 9 = 45$ )

This can be rearranged to form (first term + last term) times (number of terms divided by 2)

Therefore, since each term of a pentagonal number is $3n-2$ , a pentagonal number is in the form of $(1 + 3n - 2) \times \frac {n}{2} =$

$=\frac {n(3n-1)}{2}$

Checking, we prove by induction

Base case $n=1$ : $\frac {1 \times 2}{2} = 1$ . Therefore n=1 is true

the last term on the LHS is $3n-2$ . Therefore the next term is $3(n+1)-2 = 3n + 1$ .

Using the inductive hypothesis, we add $3n + 1$ to the RHS

$=\frac {n(3n-1)}{2} + 3n + 1$

= $\frac {n(3n-1)}{2} + \frac {6n+2}{2} = \frac {3n^2-n + 6n+2}{2} = \frac {3n^2+5n+2}{2}$

Using Null Factor Law, $n=-1$ makes this expression $=0$ , so we can factorise it with one of it's factors being $n+1$

= $\frac {(n+1)(3n+2)}{2} = \frac {(n+1)[3(n+1)-1]}{2}$

Since n + 1 is true, and n = 1 is true, this formula works.

Solving,

$\frac {n(3n-1)}{2} = 1001$

$3n^2 - n - 2002 = 0$

$n = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$n = \frac{1 \pm \sqrt{1+24024}}{6}$

$n = \frac {1 \pm 155}{6}$

$=\frac {156}{6} = \boxed{26}$ (The other solution $-154/6$ is negative and not needed)

It can be seen that

• term 1 -> 1+0+0
• term 2 -> 2+1+1
• term 3 -> 3+2+2
• term 4 -> 4+3+3
• term 5 -> 5+4+4
• term 6 -> 6+5+5 So

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using System.IO;
using System;

class Program
{
    static void Main()
    {
        int no_points =  0;
        int counter  = 1;
        while(no_points <= 1000)
        {
            no_points += counter + ( counter - 1 ) * 2;
            Console.WriteLine(no_points + " term ->" + counter);
            counter++;
        }
    }
}

Pattern of 1, 5, 12, 22, 35, 51 -> Formula of sequence is (n*(3n-1))/2. Therefore, 3n^2-n-2002 = 1001, n = 26.

I did this by figuring out how much each pentagonal number increases each time. Assuming the first picture is n = 1

Every time, 3 new lines are added of size n, but because of corners, 2 are counted twice.

Therefore, we can say that n * 3 - 2 dots are added each time. (Ex. For the first case, 1 * 3 -2 = 1 dot is added to 0)

However, n* 3 - 2 is neither a geometric nor arithmetic sequence. However, by splitting it up, you can make it both. In other words, counting the total number of ns given by each part of the equation. (pretty proud of thinking of this) Now the problem becomes easy. First, the -2 term. At any n, the - 2 part of the equation will have taken 2n dots total.

For the 3x, it gets a bit more complex, because you can't just multiply by x, so instead, find the average of all terms. A good way of doing this is adding the first and last term, than dividing by 2. (3+3x)/2. Finally, multiply again by x (average * number of terms = total), and add it back to the other formula and you'll get

(3x+3x^2)/2-2x; or x(1.5x-3.5) if you like it that way.

Now it's simply a matter of plugging in integers (which is what I did) or setting it equal to 1001 and using algebra (which I guess is the 'smart' way), and you'll find that, for x = 26, y = 1001.

p(n) = p(n-1) + 3(n-1) + 1

=> p(n) = (3n^2 - n)/2

(3n^2 - n)/2 = 1001 (quadratic equation solved)

n = 26

x is pentagonal number if and only if

there is a positive integer n with satisfies (3n^2 - n)/2 = x

The sequence proceeds as 1, 5, 12, 22, 35, between each term the differences proceed as 4, 7, 10, 13 and the difference is permanently 3, so we can deduce this must be based around square numbers. so we take 1^2 + 0 = 1, 2^2 +1 = 5, 3^2 +3 = 12, as you can see the term added is a triangle number, so the formula we use for this is ((n-1)^2+(n-1))/2 , so the formula we get for pentagonal numbers is n^2 +((n-1)^2 +(n-1))/2, which is simplified to n ( 3n -1 ) /2. Then we can form the equation n ( 3n -1 ) /2 = 1001, then you solve the equation to get n = 26

i come to two different methods to solve this: the pentagonal number of n equals either \sum {n}^2n-1­ or n^2 + \sum {i=1}^n-1

we know that any sum from 1 to any value equals half the square of n plus half the value of n, allowing us to get the formula

P= n^2 + ( (n-1)^2 + n-1) / 2

or

P=n^2 + n*(n-1)/2

therefore P= 3(n^2)/2 - n/2

Solved exactly like Prasun Biswas, except that I took P(n) - P(n-1) = 3n-2

P(n) be the 'n'th pentagon number then its easily noticed that P(n) = (3n-2) + P(n-1)