Can you count all these oranges?

Probability Level 2

The oranges are arranged such that there is 1 top orange; the second top layer has 2 more oranges than the top; the third has 3 more oranges than the second, and so on. Forming a tetrahedron of oranges, these "tetrahedral" numbers of oranges run as a series, as shown above.

What is the value of the 10 0 th 100^{\text{th}} term of this series?


The answer is 171700.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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4 solutions

When we separate this sequence by parts or layers of the oranges, we will see a pattern as shown:

a 1 a_{1} = 1

a 2 a_{2} = 1 + 3 = 1 + (1 + 2)

a 3 a_{3} = 1 + 3 + 6 = 1 + (1 + 2) + (1 + 2 + 3)

...

a n a_{n} = 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + n)

So we can see that this series is, in fact, a sum of sums of all natural numbers under n.

That is, a n a_{n} = ∑ ( ∑ n \sum (\sum n ) = ∑ ( n ( n + 1 ) 2 \sum (\frac{n(n+1)}{2} ) as we know that the sum of 1 to n equals to n ( n + 1 ) 2 \frac{n(n+1)}{2} from the formula.

Now a n a_{n} = ∑ ( n ( n + 1 ) 2 \sum (\frac{n(n+1)}{2} ) = 1 2 \frac{1}{2} [ ∑ n 2 \sum n^2 + ∑ n \sum n ]

From the formulas, ∑ n 2 \sum n^2 = n ( n + 1 ) ( 2 n + 1 ) 6 \frac{n(n+1)(2n+1)}{6} and again ∑ n \sum n = n ( n + 1 ) 2 \frac{n(n+1)}{2} , we will get:

a n a_{n} = 1 2 \frac{1}{2} [ ∑ n 2 \sum n^2 + ∑ n \sum n ] = 1 2 \frac{1}{2} [ n ( n + 1 ) ( 2 n + 1 ) 6 \frac{n(n+1)(2n+1)}{6} + n ( n + 1 ) 2 \frac{n(n+1)}{2} ]

= 1 2 \frac{1}{2} [ n ( n + 1 ) ( 2 n + 1 + 3 ) 6 \frac{n(n+1)(2n+1+3)}{6} ] = 1 2 \frac{1}{2} [ n ( n + 1 ) ( 2 n + 4 ) 6 \frac{n(n+1)(2n+4)}{6} ] = n ( n + 1 ) ( n + 2 ) 6 \frac{n(n+1)(n+2)}{6}

Therefore, a 100 a_{100} = 100 ( 100 + 1 ) ( 100 + 2 ) 6 \frac{100(100+1)(100+2)}{6} = 171,700.

As a result, there will be 171,700 oranges when 100 layers of oranges are made.

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Or you can just apply Hockey Stick Identity : ∑ n = 1 100 ( n + 1 2 ) = ( 100 + 2 3 ) = 171700 . \sum_{n=1}^{100} \dbinom{n+1}2 = \dbinom{100 + 2}3 = \boxed{171700}.

Pi Han Goh - 5 years, 6 months ago

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Oh, there's a name for this. I didn't know. Lol...

Worranat Pakornrat - 5 years, 6 months ago

But according to the question, shouldn't the 2nd layer have 3 oranges??

Yuki Kuriyama - 5 years, 6 months ago

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Yes, the second layer has 3 oranges, and the third has 6 oranges.

Worranat Pakornrat - 5 years, 6 months ago

This was my solution. But I like Pi Han Goh's more.

Let x x be our sum. Then I figured that

x = 1 ( 2 ) 2 + 2 ( 3 ) 2 + 3 ( 4 ) 2 + 4 ( 5 ) 2 + ⋯ + 99 ( 100 ) 2 + 100 ( 101 ) 2 x = \frac{1(2)}{2} + \frac{2(3)}{2} + \frac{3(4)}{2} + \frac{4(5)}{2} + \cdots + \frac{99(100)}{2}+\frac{100(101)}{2}

x = 2 ( 1 + 3 ) 2 + 4 ( 3 + 5 ) 2 + 6 ( 5 + 7 ) 2 + 8 ( 7 + 9 ) 2 + ⋯ + 98 ( 97 + 99 ) 2 + 100 ( 99 + 101 ) 2 x = \frac{2(1+3)}{2} + \frac{4(3+5)}{2} + \frac{6(5+7)}{2} + \frac{8(7+9)}{2} + \cdots + \frac{98(97+99)}{2}+\frac{100(99+101)}{2}

x = ( 1 + 3 ) + 2 ( 3 + 5 ) + 3 ( 5 + 7 ) + 4 ( 7 + 9 ) + ⋯ + 49 ( 97 + 99 ) + 50 ( 99 + 101 ) x = (1+3) + 2(3+5) + 3(5+7) + 4(7+9) + \cdots + 49(97+99)+ 50(99+101)

x = 1 ⋅ 4 + 2 ⋅ 8 + 3 ⋅ 12 + 4 ⋅ 16 + ⋯ + 49 ⋅ 196 + 50 ⋅ 200 x = 1 \cdot 4 + 2 \cdot 8 + 3 \cdot 12 + 4 \cdot 16 + \cdots + 49 \cdot 196+ 50 \cdot 200

x = 4 ⋅ [ 1 2 + 2 2 + 3 2 + 4 2 + ⋯ + 4 9 2 + 5 0 2 ] x = 4 \cdot [1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + 49^{2}+ 50^{2}]

x = 4 ⋅ [ 50 ⋅ 51 ⋅ 101 6 ] x = 4 \cdot [\frac{50 \cdot 51 \cdot 101}{6}]

x = 4 ⋅ [ 50 ⋅ 51 ⋅ 101 6 ] x = 4 \cdot [\frac{50 \cdot 51 \cdot 101}{6}]

x = 171700 x = \boxed{171700}

Patrick Engelmann - 5 years, 6 months ago
Andy Hayes
Nov 7, 2016

Each layer of oranges is a triangular number. Triangular numbers can be computed with the binomial coefficient.

∑ k = 1 n k = ( n + 1 2 ) \sum\limits_{k=1}^{n}{k}=\binom{n+1}{2}

The sum of triangular numbers can be computed with the hockey stick identity .

∑ k = 1 n ∑ j = 1 k j = ∑ k = 1 n ( k + 1 2 ) = ( n + 2 3 ) \sum\limits_{k=1}^{n}\sum\limits_{j=1}^{k}{j}=\sum\limits_{k=1}^{n}\binom{k+1}{2}=\binom{n+2}{3}

So the sum of the first 100 triangular numbers is:

∑ k = 1 100 ∑ j = 1 k j = ( 102 3 ) = 171700 \sum\limits_{k=1}^{100}\sum\limits_{j=1}^{k}{j}=\binom{102}{3}=\boxed{171700}

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Ariella Lee
Dec 9, 2015

I counted layer by layer, just like Worranat Pakornrat, but calculated the sum a different (sort of) way.

The top layer has 1 1 orange. The next layer as 1 1 orange in its first row, then 2 2 oranges in its second row, so 1 + 2 1+2 oranges. The third layer has 1 + 2 + 3 1+2+3 oranges. And so on. The 100 100 th layer will have 1 + 2 + ⋯ + 100 1+2+\cdots+100 oranges.

So we want to compute: 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ⋯ + ( 1 + 2 + ⋯ + 100 ) 1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+100) . There are 100 100 ones, 99 99 twos, 98 98 threes, and so on, so we have: ( 1 ⋅ 100 ) + ( 2 ⋅ 99 ) + ( 3 ⋅ 98 ) + ⋯ + ( 100 ⋅ 1 ) (1\cdot100)+(2\cdot99)+(3\cdot98)+\cdots+(100\cdot1) . This can be represented by ∑ k = 1 100 k ( 101 − k ) \sum_{k=1}^{100}k(101-k) , which I used the computer calculate, or you can write it as 101 ∑ k = 1 100 k − ∑ k = 1 100 k 2 101\sum_{k=1}^{100}k-\sum_{k=1}^{100}k^{2} and use the formulas.

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Nikola Djuric
Dec 14, 2015

S=1+(1+2)+(1+2+3)+...+(1+...+100)= 100×1+(100-1)×2+...(100-99)×100= 100×(1+2+...+100)-1x2-2x3-..-99×100= 100×100×101/2-2(1+(1+2)+(1+2+3)+ ...+(1+2+3+...+99))= 100²×101/2-2(S-100×101/2) So 3S=100²×101/2+100×101= 100×101×(100+2)/2 So S=100×101×102/6=171700

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