summation prob
- Thread starter sujoy
- Start date
If you write the sum of each bracketed term you get 1 + 3 + 6 + 10 + 15 +...n, the triangular numbers.
The sequence of triangular numbers form a finite difference sequence of the form an^2 + bn + c with 2nd differences being constant at 1.
n.......1......3......6......10......15...
Diff.......2......3........4.......5
Diff...........1.......1........1
Using ths data, we can write
a(1)2 + b(1) + c = 1
a(2^2) + b(2) + c = 3
a(3^2) + b(3) + c = 6
Solving, a 1/2, b = 1/2 and c = 0 giving us the sum of the triangular numbers as St = (n^2 + n)/2.
The nth term is (1 + 3 + 6 + 10 + 15 +....n)
The sequence of triangular numbers form a finite difference sequence of the form an^2 + bn + c with 2nd differences being constant at 1.
n.......1......3......6......10......15...
Diff.......2......3........4.......5
Diff...........1.......1........1
Using ths data, we can write
a(1)2 + b(1) + c = 1
a(2^2) + b(2) + c = 3
a(3^2) + b(3) + c = 6
Solving, a 1/2, b = 1/2 and c = 0 giving us the sum of the triangular numbers as St = (n^2 + n)/2.
The nth term is (1 + 3 + 6 + 10 + 15 +....n)
Hello, sujoy!
. <sub>n</sub> . . . . . . . . . . . . . . . . <sub>n</sub>
. ∑ k .= .n(n + 1)/2 . . . ∑ k<sup>2</sup> .= .n(n + 1)(2n + 1)/6
<sup>k=1</sup> . . . . . . . . . . . . . . <sup>k=1</sup>
As TchrWill pointed out, we are adding Triangular Numbers of the form: k(k+1)/2
. . . . . . . . . . . .<sub>n</sub> . . . . . . . . . . . . . . . <sub>n</sub> . . . . . . . . .<sub>n</sub>
So we have: . ∑ k(k + 1)/2 .= .(1/2)∑ k<sup>2</sup> + (1/2) ∑ k
. . . . . . . . . . .<sup>k=1</sup> . . . . . . . . . . . . . <sup>k=1</sup> . . . . . . . <sup>k=1</sup>
Applying the formulas: . (1/2)[n(n + 1)/2] + (1/2)[n(n + 1)(2n + 1)/6]
. . which simplifies to: . n(n + 1)(n + 2)/6
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I worked on this problem years ago when a friend had asked:
. . "How many gifts were given during The Twelve Days of Christmas?"
Are you familiar with these summation formulas?
. <sub>n</sub> . . . . . . . . . . . . . . . . <sub>n</sub>
. ∑ k .= .n(n + 1)/2 . . . ∑ k<sup>2</sup> .= .n(n + 1)(2n + 1)/6
<sup>k=1</sup> . . . . . . . . . . . . . . <sup>k=1</sup>
As TchrWill pointed out, we are adding Triangular Numbers of the form: k(k+1)/2
. . . . . . . . . . . .<sub>n</sub> . . . . . . . . . . . . . . . <sub>n</sub> . . . . . . . . .<sub>n</sub>
So we have: . ∑ k(k + 1)/2 .= .(1/2)∑ k<sup>2</sup> + (1/2) ∑ k
. . . . . . . . . . .<sup>k=1</sup> . . . . . . . . . . . . . <sup>k=1</sup> . . . . . . . <sup>k=1</sup>
Applying the formulas: . (1/2)[n(n + 1)/2] + (1/2)[n(n + 1)(2n + 1)/6]
. . which simplifies to: . n(n + 1)(n + 2)/6
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I worked on this problem years ago when a friend had asked:
. . "How many gifts were given during The Twelve Days of Christmas?"
OOPS
I got called away for something and submitted the previous reply before finishing it.
Can we add:
1+[1+2] +[1+2+3]+ [1+2+3+4]..................n terms
where the last term is (1+2+3+4+......+n)
--------------------------------------------------------------------------------
If you write the sum of each bracketed term you get 1 + 3 + 6 + 10 + 15 +...n, the triangular numbers.
The sequence of triangular numbers form a finite difference sequence of the form an^2 + bn + c with 2nd differences being constant at 1.
n.......1......3......6......10......15...
Diff.......2......3........4.......5
Diff...........1.......1........1
Using ths data, we can write
a(1)2 + b(1) + c = 1
a(2^2) + b(2) + c = 3
a(3^2) + b(3) + c = 6
Solving, a 1/2, b = 1/2 and c = 0 giving us the sum of the triangular numbers as St = (n^2 + n)/2.
The nth term is (1 + 3 + 6 + 10 + 15 +....n)
Now if you sum these terms, you get
n......1......2......3......4......5...
Sn....1......4.....10....20....35....n not surprisingly, another finite difference sequece as in
n......1......2......3......4......5
S......1......4.....10....20....35
Diff.......3......6....10....15
Diff...........3.....4......5
Diff..............1......1
With the third differences being constant, the sequences of sums is a finite difference sequence of the 3rd order making the general expression for the nth term S = an^3 + bn^2 + cn + d.
Using the data, we can write
a(1^3) + b(1^2) + c(1) + d = 1
a(2^3) + b(2^2) + c(2) + d = 4
a(3^3) + b(3^2) + c(3) + d = 10
a(4^3) + b(4^2) + c(4) + d = 20
Solving, a = 1/6, b = 1/2, c = 1/3 and d = 0.
Therefore, the nth sum of the n sums is given by Sn = (n^3 + 3n^2 + 2n)/6, the same expression that soroban so elagantly presented to you earlier.
Sorry if it caused you any confusion.
I got called away for something and submitted the previous reply before finishing it.
Can we add:
1+[1+2] +[1+2+3]+ [1+2+3+4]..................n terms
where the last term is (1+2+3+4+......+n)
--------------------------------------------------------------------------------
If you write the sum of each bracketed term you get 1 + 3 + 6 + 10 + 15 +...n, the triangular numbers.
The sequence of triangular numbers form a finite difference sequence of the form an^2 + bn + c with 2nd differences being constant at 1.
n.......1......3......6......10......15...
Diff.......2......3........4.......5
Diff...........1.......1........1
Using ths data, we can write
a(1)2 + b(1) + c = 1
a(2^2) + b(2) + c = 3
a(3^2) + b(3) + c = 6
Solving, a 1/2, b = 1/2 and c = 0 giving us the sum of the triangular numbers as St = (n^2 + n)/2.
The nth term is (1 + 3 + 6 + 10 + 15 +....n)
Now if you sum these terms, you get
n......1......2......3......4......5...
Sn....1......4.....10....20....35....n not surprisingly, another finite difference sequece as in
n......1......2......3......4......5
S......1......4.....10....20....35
Diff.......3......6....10....15
Diff...........3.....4......5
Diff..............1......1
With the third differences being constant, the sequences of sums is a finite difference sequence of the 3rd order making the general expression for the nth term S = an^3 + bn^2 + cn + d.
Using the data, we can write
a(1^3) + b(1^2) + c(1) + d = 1
a(2^3) + b(2^2) + c(2) + d = 4
a(3^3) + b(3^2) + c(3) + d = 10
a(4^3) + b(4^2) + c(4) + d = 20
Solving, a = 1/6, b = 1/2, c = 1/3 and d = 0.
Therefore, the nth sum of the n sums is given by Sn = (n^3 + 3n^2 + 2n)/6, the same expression that soroban so elagantly presented to you earlier.
Sorry if it caused you any confusion.
