How do you write this in Summation Notation, 2+6+11+17+24?

Question

I need it for our statistic assignment, please. Thank you

Answer 1

In order to put something in summation notation, we need to find a formula or pattern for the what the terms look like, based on what number term they are!

 

Do you see a pattern to these terms in the sum??

 

Lets first consider a sequence: 2,6,11,17,24

 

It appears as though the numbers are all increasing, and in fact the differences between successive numbers is increasing too!

 

6-2=4

11-6=5

17-11=6

24-17=7

 

In fact this difference is increasing by 1 every time!

 

So lets call the first term in the sequence a_0 = 2

 

What we notice is that

a_1 = a_0+4

a_2 = a_1+5

a_3=a_2+6

a_4=a_3+7

 

We would like to find a formula to express a_n as a function of n only and not any of the other terms of the sequence. Lets first write out each term, in terms of a_0

 

a_0 = 2 = a_0

a_1 = 6 = a_0 + 4

a_2 =11= a_1+5 = a_0 +4+5

a_3 =17= a_2+6 = a_0 +4+5+6

a_4 =24= a_3+7 = a_0 +4+5+6+7

 

Do you see a pattern for the a_n's??

 

a_n = a_0 + SUM_{i=1...n} (i+3)       for every n>0

 

Now that we have a formula for the terms of the sequence {a_n}_0^4  meaning {a_0,a_1,...,a_4}, the SUMMATION given as your problem is the sum of these terms of the sequence...so we can use a second summation notation, to sum up these a_n's

 

2+6+11+17+24 = SUM_{n=0...4} [ 2+SUM_{i=1...n} (i+3) ] 

 

Where it is understood that the inner sum from i=1 to n=0 is 0.

 

Since Gauss has shown us a way to collapse the formula for the inner-sum, the recurrence relation described above has a closed form without a sum, and we can write this a little more neatly.

 

What is SUM_{i=1...n} i ?

This is just adding the first n whole numbers together, as legend says Gauss did very quickly.... = n(n+1)/2

Pair the numbers by writing a second copy of them backwards and add pairwise to get n(n+1), then divide by 2 for the duplicate copy.

 

Thus SUM{i=1...n} (i+3) = n(n+1)/2 + SUM_{i=1...n} 3 = n(n+1)/2 + 3n

 

So we can collapse the inner sum, or more precisely give a closed formula for a_n without summation notation:

 

a_n = 2 + n(n+1)/2 + 3n

 

Lets check that this outputs the numbers in our sequence/series:

 

a_0 = 2 + 0 + 0

a_1 = 2 + 1 + 3 = 6

a_2 = 2 + 3 + 6 = 11

a_3 = 2 + 6 + 9 = 17

a_4 = 2 + 10 + 12 = 24

 

Looks like we are on the right track!

 

So the series or summation you have given can be expressed in summation notation with a single sum, as follows:

 

2+6+11+17+24 = SUM_{n=0...4}  (2 + n(n+1)/2 + 3n)

 

Hope that helps!