Sequence and series

Question

If

n     i     j
Σ    Σ    Σ 1= 220
i=1 j=1 k=1

 

Then find the value of n

Answer 1

We can break this down to quickly writing down the first few terms in three sequences.

 

Let A_j = ∑_k=1,...,j 1 = j, B_j = ∑_j=1,...,i A_j, C_n = ∑_I=1,...,n B_i.

 

Then we want an n such that C_n = 220.

 

A's: 1,  2,   3,  4,   5,   6,   7,   8,     9,   10, ...

B's: 1,  3,   6, 10, 15, 21, 28,  36,   45,  55, ...

C's: 1,  4, 10, 20, 35, 56, 84, 120, 165, 220, ...

 

The we see that C₁₀ = 220, so n=10.

Answer 2

Let's say n = 3.

 

So i goes from 1 to 3, which means j and k will do the same (sort of).  Just need to be very careful as you go.

 

When i = 1, the left summation has 1 term, which is when j = 1, which is then when k = 1.  Carrying out the right summation, you get an overall first term of 1.  (Pretty boring.)

 

When i = 2, j will go from 1 to 2, which means the right summation will be carried out twice.  Once from 1 to 1 (as above), and then once from 1 to 2.  This means that the overall sum is 1 + 2 = 3.

 

When i = 3, j goes from 1 to 3, right summation happens 3 times, which means we'll end up with 1 + 2 + 3 = 6 as our 3rd overall term.

 

So when n = 3, our overall sum is just 10.  Luckily it doesn't take too long to get to 220 if you need to show it all out by hand.  But the pattern continues like this where the ith term is just 1 + 2 + ... + i.  These are also known as the triangular numbers if you want to look up the sequence.

 

Hope that makes sense!