Computer Science Canada

the sum of the sum of n integral squares
Author:  batman12 [ Fri Jan 06, 2006 1:30 pm ]
Post subject:  the sum of the sum of n integral squares
hi ,
i am using the this forumla for the sum of n integreal number.
i am trying to find the sum by using this 1**2+2**2+3**2+4**2+5**2.....n**2=x
where the user enters the n value.
please give me suggestions how to do this in turing. i am stuck.anyhelp is aprreicated. thank you.thanks.
Author:  iker [ Fri Jan 06, 2006 7:32 pm ]
Post subject: 
use for loop such as
code:

var n, x : int := 0
put "enter the ending value"
get n
for i : 1 .. n
    x := x + 2 ** i
end for
put x
Author:  Cervantes [ Fri Jan 06, 2006 8:12 pm ]
Post subject: 
iker: Look at your for loop closely.
code:

for i : 1 .. n
    x := x + 2 ** i
end for

2 ** i is much different than i ** 2


A better way to solve this problem is to use the formula for the sum of the squares of the natural numbers, which is n(n + 1)(2n + 1) / 6

Allow me to demonstrate:
code:

fcn sum_of_natural_squares (n : int) : int
    result round (n * (n + 1) * (2 * n + 1) / 6)
end sum_of_natural_squares

fcn sum_of_natural_squares_2 (n : int) : int
    var final := 0
    for i : 1 .. n
        final += i ** 2
    end for
    result final
end sum_of_natural_squares_2

for i : 1 .. maxrow - 1

    put i : 4, sum_of_natural_squares (i) : 10, sum_of_natural_squares_2 (i) : 10

end for
Author:  iker [ Fri Jan 06, 2006 11:59 pm ]
Post subject: 
Cervantes wrote:
iker: Look at your for loop closely.
code:

for i : 1 .. n
    x := x + 2 ** i
end for

2 ** i is much different than i ** 2

Embarassed thats embarassing, the one thing I noticed is that your code is basically the same as my code, but expressed in a different way
also instead of having
code:

for i : 1 .. n
    x := x + i ** 2
end for

It could have
code:

for i : 1 .. n
    x += i ** 2
end for

which saves about 3 characters...
but it doesn't realy matter...
Author:  batman12 [ Sat Jan 07, 2006 11:40 am ]
Post subject:  thnaks very much iker and cervantes for you help
hey guys,
thanks for the help. i got it. thanks for the suggestions. it really helped me out. thanks. Very Happy Very Happy Very Happy Very Happy Very Happy Very Happy Very Happy
Author:  person [ Sat Jan 07, 2006 4:52 pm ]
Post subject: 
Quote:
n(n + 1)(2n + 1) / 6


can u post a proof of y this works, or is this one of those math thngs thats not currently proovable?
Author:  Avarita [ Sat Jan 07, 2006 6:05 pm ]
Post subject: 
ok... let's prove it.

The first 5 terms (with the 1st term at 0) of the sequence are:
0, 1, 5, 14, 30

first difference (the difference between each term) = 1, 4, 9, 16 ...

second difference (the difference between each term of the first difference) = 3, 6, 9

third difference (the difference between each term of the second difference) = 3, 3, 3

Because the third difference terms are all equal (ie, all being 3), the relationship is cubic, and the standard form for a cubic equation is...

y = ax**3 + bx**2 + cx + d

Subbing in x = 0 and y = 0 gives us d = 0.

Subbing in (1, 1), (2, 5), (3, 14) gives us the following:
1. a + b + c = 1 -> a = 1 - b - c
2. 8a + 4b + 2c = 5
3. 27a + 9b + 3c = 14

Subbing equation #1 in both #2 and #3 gives us the following 2 equations:
4. 8(1 - b - c) + 4b + 2c = 5 -> 4b + 6c = 3
5. 27(1 - b - c) + 9b + 3c = 14 -> 18b + 24b = 13

Multipling equation #4 by 4 then subtract from equation #5:
5. 2b = 1 -> b = 1/2

So now sub b = 1/2 into equation #4:
6. 4(1/2) + 6c = 3 -> c = 1/6

Sub c = 1/6 and b = 1/2 into equation #1:
7. a = 1 - 1/2 - 1/6 -> a = 1/3

So now we have this equation:
y = (x**3) / 3 + (x**2) / 2 + x/6

Factor out x/6 and we have:
y = x/6 (2x**2 + 3x + 1)

Factor this equation and we have:
y = x/6 (2x + 1)(x + 1)

Which is the same as f(n) = n(n + 1)(2n + 1) / 6

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