Hey! I have seen that before

Algebra Level 4

$\large 2,6,12,20,30, \ldots, a_n$

Above shows the first few numbers of an increasing series. If the sum of these numbers equals 32430, find the value of the last term, $a_n$ .

The answer is 2070.

6 solutions

Ikkyu San
May 1, 2015

$\begin{aligned}2+6+12+20+\ldots+a_{n}=&\ 32430\\(1^2+\color{#D61F06}1)+(2^2+\color{#D61F06}2)+(3^2+\color{#D61F06}3)+(4^2+\color{#D61F06}4)+\ldots+(n^2+\color{#D61F06}n)=&\ 32430\\\displaystyle\sum_{i=1}^{n}(i^2+\color{#D61F06}i)=&\ 32430\\\displaystyle\sum_{i=1}^{n}i^2+\color{#D61F06}{\displaystyle\sum_{i=1}^{n}i}=&\ 32430\\\dfrac{n(n+1)(2n+1)}{6}+\color{#D61F06}{\dfrac{n(n+1)}{2}}=&\ 32430\\\dfrac{2n^3+3n^2+n+3n^2+3n}{6}=&\ 32430\\\dfrac{n^3+3n^2+2n}{3}=&\ 32430\\n^3+3n^2+2n-97290=&\ 0\\(n-45)(n^2+48n+2162)=&\ 0\\n=&\ 45\end{aligned}$

From sequence $2,6,12,20,30,...\rightarrow 1^2+1,2^2+2,3^2+3,4^2+4,5^2+5,...$

Therefore, $a_n=n^2+n\rightarrow a_{45}=45^2+45=2025+45=\boxed{2070}$

Moderator note:

Your work is very neat, I like that you incorporate colours into your work. Great job!

Bonus question: Instead of triangular numbers, we are given the sum of the first $n$ smallest pentagonal numbers is 4200. Can you find the value of $n$ ?

Challenge Master : Thank you! // The value of n is 20.

Ikkyu San - 6 years, 1 month ago

Challenge Master: The value of $\text{n}$ is 20, then!

Kartik Sharma - 6 years, 1 month ago

Yeah! A good solution

A Former Brilliant Member - 6 years, 1 month ago

Thank you very much!

Ikkyu San - 6 years, 1 month ago

How you do that Sir? The 2nd question is harder i thought.

Hafizh Ahsan Permana - 6 years, 1 month ago

General sequence of pentagonal number is $1,5,12,22,35,\ldots$

And general term of pentagonal number is $p_n=\dfrac{3n^2-n}{2}$

For sum of pentagonal number is,

$\begin{aligned}1+5+12+22+\ldots+p_{\color{#D61F06}n}=&\ 4200\\\dfrac{3(\color{#D61F06}1^2)-\color{#D61F06}1}{2}+\dfrac{3(\color{#D61F06}2^2)-\color{#D61F06}2}{2}+\dfrac{3(\color{#D61F06}3^2)-\color{#D61F06}3}{2}+\dfrac{3(\color{#D61F06}4^2)-\color{#D61F06}4}{2}+\ldots+\dfrac{3\color{#D61F06}n^2-\color{#D61F06}n}{2}=&\ 4200\\\displaystyle\sum_{i=1}^{\color{#D61F06}n}\left(\dfrac{3i^2-i}{2}\right)=&\ 4200\\\dfrac{3}{2}\displaystyle\sum_{i=1}^{\color{#D61F06}n}i^2-\dfrac{1}{2}\displaystyle\sum_{i=1}^{\color{#D61F06}n}i=&\ 4200\\\left(\dfrac{3}{2}\right)\left(\dfrac{\color{#D61F06}n(\color{#D61F06}n+1)(2\color{#D61F06}n+1)}{6}\right)-\left(\dfrac{1}{2}\right)\left(\dfrac{\color{#D61F06}n(\color{#D61F06}n+1)}{2}\right)=&\ 4200\\\dfrac{2\color{#D61F06}n^3+3\color{#D61F06}n^2+\color{#D61F06}n}{4}-\dfrac{\color{#D61F06}n^2+\color{#D61F06}n}{4}=&\ 4200\\\dfrac{\color{#D61F06}n^3+\color{#D61F06}n^2}{2}=&\ 4200\\\color{#D61F06}n^3+\color{#D61F06}n^2-8400=&\ 0\\(\color{#D61F06}n-20)(\color{#D61F06}n^2+21\color{#D61F06}n+420)=&\ 0\\\color{#D61F06}{n=}&\ \color{#D61F06}{20}\end{aligned}$

Ikkyu San - 6 years, 1 month ago

Challenge Master: the value of n is 20

Krishna Ramesh - 6 years, 1 month ago

That's right the answer for this bonus question is 20. Good work @Krishna Ramesh , @Ikkyu San , @Kartik Sharma !

Brilliant Mathematics Staff - 6 years, 1 month ago

Yes, we will get n n (n+1) = 8400. This gives n = 20

Raushan Sharma - 6 years, 1 month ago

An alternate method: Notice that $a_{n+3}-3a_{n+2}+3a_{n+1}-a_n=0$ for any $n$ . (The binomial coefficients here are not a coincidence! I think that the easiest way to prove it is to just bash the algebra, though...) Using this, we could sum this similarly to a geometric series: Call the sum $S$ . Write down $S$ , $-3S$ , $3S$ , and (S) in such a way to use the above identity, and... well, you'll see, I'm fairly certain it all works out in the end.

Akiva Weinberger - 6 years, 1 month ago

On that identity: It's weird.

If $a_n$ is constant, then we have $a_{n+1}-a_n=0$ .

If $a_n$ is linear, then we have $a_{n+2}-2a_{n+1}+a_n=0$ .

If $a_n$ is quadratic, then we have $a_{n+3}-3a_{n+2}+3a_{n+1}-a_n=0$ .

If $a_n$ is cubic, then we have $a_{n+4}-4a_{n+3}+6a_{n+2}-4a_{n+1}+a_n=0$ .

Notice the similarities to the polynomials $(a-1)^1,(a-1)^2,(a-1)^3,\dots$ . The pattern continues, but I'm not sure I know the simplest way to prove this.

Akiva Weinberger - 6 years, 1 month ago

Nice observation! Sketch out the pascal triangle, and you can see why.

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – After writing that, I realized it could be proven easily by induction.

(I wonder if it works for fractional orders, similar to Newton's generalized binomial theorem?)

Akiva Weinberger - 6 years, 1 month ago

@Akiva Weinberger – POST PROOF HERE! Fractional orders too? how?

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh – Wait, I don't know how to prove it for fractional orders (EDIT: "degrees"). I just wondered if it could be true.

For the induction thing, just notice that $f(n+1)-f(n)$ brings the degree of $f$ down one. Applying this transformation to $a_n$ repeatedly, we get the identities above.

Akiva Weinberger - 6 years, 1 month ago

Great solution.

Chaitnya Shrivastava - 5 years, 7 months ago

Vishnu Bhagyanath
May 5, 2015

$2+6+12+20+30+ \ldots+a_n = 32430$ $2(1+3+6+10+15+ \ldots+\frac{a_n}{2}) = 32430$ Using the sum for triangular series formula, $2(\frac{(n)(n+1)(n+2)}{6}) = 32430$ $(n)(n+1)(n+2) = 97290$ $\implies n = 45$ Since $a_n = n(n+1)$ $a_n = 45(46)$ $\boxed{a_n = 2070}$

Moderator note:

Yes, a slight variation. To make this solution clearer, you could convert the triangular numbers in terms of binomial coefficients. Nevertheless, great work!

A Tetrahedral number is the sum of the first triangular numbers. One could easily obtain its identity by the hockey stick identity: $\large {2 \choose 2} + {3 \choose 2} + {4 \choose 2} + \ldots + {n+1 \choose 2} = {n+2 \choose 3}$

Niranjan Khanderia
May 2, 2015

Just Ikkyu San method with little modification.
$\displaystyle \sum_{i=1}^n(i^2+i)\\=\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}\\= \dfrac{n(n+1)}{2}* \dfrac{2n+4}{3}\\=\dfrac{n(n+1)(n+2)}{3}\\=32430.\\\therefore ~\dfrac{(n+1)^3} 3\\ \approx 32430.\\~~~~~~~~~\implies~n=44.99.....\\~~~~~~~~~~~~Take~n=45,\\\dfrac{45*46*47} 3 \\=32430.\\ ~~~~~~~~~~~\implies~n=45, \\a_n\\=45^2+45\\=2070.$

Moderator note:

Wow, a quick way to force out a solution. Nicely done! One should note that $\frac {n(n+1)(n+2)}{3}$ is an increasing function, so there's only one unique value when it's positive.

Nicola Hu
May 5, 2015

The sum of the sequence is : $2+6+12+....=2(1+(1+2)+(1+2+3)+....+(1+2+...n))$

The problem is : how much is $2S$ Let's find S $S=( 1 + (1+2) + (1+2+3) + ......+ (1+2+....n))$

Let' draw a picture!

$Q + S = \tfrac{n(n+1)^2}{2}$ $Q= 1^2+2^2+3^2+...+n^2=\tfrac{n(n+1)(2n+1)}{6}$ And so $S=(Q+S)-Q=\tfrac{n(n+1)^2}{2} -\tfrac{n(n+1)(2n+1)}{6} = \tfrac{n(n+1)(n+2)}{6}$

From this , we obtain $2S=\tfrac{n(n+1)(n+2)}{3}=32340$ Notice that $n(n+1)(n+2) \sim n^3 .$ $\sqrt[3]{32340*3} \sim 45.$ We try $n=44$ and it works

Then $a_n= n(n+1)= 2070$

David Holcer
May 2, 2015

For me, it's easier to see a pattern with recursive loops, so using a bit of python to help me. Although, it is much easier to do Ikkyu's method if you notice the pattern.

x=[2]
c=2
while True:
    d=x[-1]+2*c
    if sum(x)==32430:
        break
    x.append(d)
    c+=1
print x[-1]

Devendra Kumar Singh
Aug 30, 2015

Hey! I have seen that before

The answer is 2070.

6 solutions

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