Sum of First Positive Squared Numbers

I have been practicing with the difference analysis technique because it seems to me a very good tool, which has many benefits. We practiced very little of the technique in class, so I asked Antonio to help me, and put me an example. He told me that the problem got harder depending on te quantity of levels, so he told me to try a 3rd level problem.

My goal was to reach a formula for the sum of the first positive squared integers.

Here is my work:



First I listed the first terms

1, 4, 9, 16, 25 - But since what I want is a formula for the sum, I created a sequence with the sum of all previous terms.

1, 5 (1+4), 14 (1+4+9), 30 (1+4+9+16), 55 (1+4+9+16+25).......



Now to find the level, I have to reach the common difference.

1, 5, 14, 30, 55..

Ist 4, 9, 16, 25

2nd 5, 7, 9

3rd 2, 2

(Now I can assure the formula is raised to the 3rd degree)





I identified the values so when x = 1, y = 1.. this means that n = x and Tn = y.

x = 1, y = 1

x = 2, y = 5

x = 3, y = 14

x = 4, y = 30



Next I plugged in the values in the formula ax^3 + bx^2 + cx + d = y, and then solve for the variables by subtraction (which is pure algebra)



a(1)^3 + b(1)^2 + c(1) + d = 1, so, a + b + c + d = 1, when x = 1

a(2)^3 + b(2)^2 + c(2) + d = 5, so 8a + 4b + 2c +d = 5, when x = 2

a(3)^3 + b(3)^2 + c(3) + d = 14, so 27a + 9b + 3c + d = 14, when x = 3

a(4)^3 + b(4)^2 + c(4) + d = 30, so, 64a + 16b + 4c + d = 30, when x = 4



Now, by subtraction solve for the variables.

8a + 4b + 2c + d = 5, minus

a + b + c + d = 1, which equals

7a + 3b + c = 4 ( We will use this later and call this formula 1 or f1, to return quicker)



Then repeat the same process with different formulas

27a + 9b + 3c + d = 14 minus

8a + 4b + 2c + d = 5, which equals

19a + 5b + c = 9, hang on there keep this one to, and call it f2



Now we can use this to formulas to eliminate one variable.

19a + 5b + c = 9 minus

7a + 3b + c = 4 which equals

12a + 2b = 5, f3



Repeat again

64a +16b + 4c +d = 30 minus

27a + 9b + 3c + d = 14 equals

37a + 7b + c = 16, f4



Now subtract f2 from f4 to eliminate c once again.

37a + 7b + c = 16 minus

19a + 5b + c = 9 equals

18a + 2b = 7, this is f5



Okay were almost finished, now subtract f3 from f5

18a + 2b = 7, minus

12a + 2b = 5 equals

6a = 2, Finally we reach our first variable a = 1/3



Now just plug in a in f5 to solve for b,

18(1/3) + 2b = 7

6 + 2b = 7

2b = 1, now we also have b = 1/2



With a and b now we can solve for c plugging in f4

37(1/3) + 7(1/2) + c = 16

(37/3) + (7/2) + c = 16 so since the Least common multiple is 6 we multiply the denominators to get 6, then we are able to subtract.

(74/6) + (21/6) + c = 16

16 - (95/6) = c

(96/6) - (95/6) = c

So, finally we have c which equals 1/6 = c

We only have one last variable to find which is d.

So, we repeat the same process. Now we can just plug in a + b + c + d + = 1, rememeber this is the first formula we used.

1/3 + 1/2 + 1/6 + d = 1, so we repeat the LCM rule and we simplify to

2/6 + 3/6 + 1/6 + d = 1

6/6 + d = 1

1 + d = 1

d = 0



We finally have all our variables defined, that was the long part so,

a = 1/3

b = 1/2

c = 1/6

d = 0



We plug all this in the normal formula ax^3 + bx^2 + cx + d = y

Also, lets replace x for n so that we now we are talking about sequences and series.

(1/3)n^3 + (1/2)n^2 + (1/6)n = Sn, where Sn is the sum to the nth term of all positive integers squared.

So lets proove if it works.

Lets try to find the sum to 6th term

S6 = (1/3)6^3 + (1/2)6^2 + (1/6)6

S6 = 72 + 18 + 1

S6 = 91



1+4+9+16+25+36 = 91, so it does work



I really think this is a very important tool for seqeunces and series, beacuse we have only been taught to find formulas for geometric and arithmetic series, as you can see this is neither, this is called a complex sequence. I think you should try this and get good at this, it can save you many times.



Example problem, try maybe an easier one to practice try to find the sum for the sequence for first positive odd integers. In another post I did it for the terms, now you do it for the sum.

The terms are:
1 , 4, 9, 16, 25....