Solving summation?
- Thread starter eritar
- Start date
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
Here is a reference that may help.
You can rewrite as \(\sum\limits_{k = 1}^n {({k^6} + 5{k^4} + 7)}\\ = \sum\limits_{k = 1}^n {({k^6}} ) + 5\sum\limits_{k = 1}^n {({k^4}) + } \sum\limits_{k = 1}^n {(7)}\\ = \sum\limits_{k = 1}^n {({k^6}} ) + 5\sum\limits_{k = 1}^n {({k^4}) + } 7n\)
In many calculus textbooks there are lists of so called special sums.
- Joined
- Nov 24, 2012
- Messages
- 3,021
What I would do is write the difference equation:
[MATH]S_{n}=S_{n-1}+n^6+5n^4+7[/MATH]
The homogeneous solution will be a constant:
[MATH]h_n=c_1[/MATH]
And the particular solution will be a 7th degree polynomial:
[MATH]p_n=An^7+Bn^6+Cn^5+Dn^4+En^3+Fn^2+Gn[/MATH]
Plugging this into the difference equation, we obtain:
[MATH]7An^6+(-21A+6B)n^5+(35A-15B+5C)n^4+(-35A+20B-10C+4D)n^3+(21A-15B+10C-6D+3E)n^2+(-7A+6B-5C+4D-3E+2F)n+(A-B+C-D+E-F+G)=n^6+5n^4+7[/MATH]
Equating corresponding coefficients, we obtain the system:
[MATH]7A=1[/MATH]
[MATH]-21A+6B=0[/MATH]
[MATH]35A-15B+5C=5[/MATH]
[MATH]-35A+20B-10C+4D=0[/MATH]
[MATH]21A-15B+10C-6D+3E=0[/MATH]
[MATH]-7A+6B-5C+4D-3E+2F=0[/MATH]
[MATH]A-B+C-D+E-F+G=7[/MATH]
We then find:
[MATH]A=\frac{1}{7}[/MATH]
[MATH]B=\frac{1}{2}[/MATH]
[MATH]C=\frac{3}{2}[/MATH]
[MATH]D=\frac{5}{2}[/MATH]
[MATH]E=\frac{3}{2}[/MATH]
[MATH]F=0[/MATH]
[MATH]G=\frac{48}{7}[/MATH]
And so our particular solution is:
[MATH]p_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n[/MATH]
And so by the principle of superposition, we find:
[MATH]S_n=p_n+h_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n+c_1[/MATH]
Now we can use the initial sum to determine the parameter:
[MATH]S_1=\frac{1}{7}+\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\frac{3}{2}+\frac{48}{7}+c_1=13\implies c_1=0[/MATH]
And so we find the solution is:
[MATH]S_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n[/MATH]
Or:
[MATH]S_n=\frac{n}{14}(2n^6+7n^5+21n^4+35n^3+21n^2+96)[/MATH]
[MATH]S_{n}=S_{n-1}+n^6+5n^4+7[/MATH]
The homogeneous solution will be a constant:
[MATH]h_n=c_1[/MATH]
And the particular solution will be a 7th degree polynomial:
[MATH]p_n=An^7+Bn^6+Cn^5+Dn^4+En^3+Fn^2+Gn[/MATH]
Plugging this into the difference equation, we obtain:
[MATH]7An^6+(-21A+6B)n^5+(35A-15B+5C)n^4+(-35A+20B-10C+4D)n^3+(21A-15B+10C-6D+3E)n^2+(-7A+6B-5C+4D-3E+2F)n+(A-B+C-D+E-F+G)=n^6+5n^4+7[/MATH]
Equating corresponding coefficients, we obtain the system:
[MATH]7A=1[/MATH]
[MATH]-21A+6B=0[/MATH]
[MATH]35A-15B+5C=5[/MATH]
[MATH]-35A+20B-10C+4D=0[/MATH]
[MATH]21A-15B+10C-6D+3E=0[/MATH]
[MATH]-7A+6B-5C+4D-3E+2F=0[/MATH]
[MATH]A-B+C-D+E-F+G=7[/MATH]
We then find:
[MATH]A=\frac{1}{7}[/MATH]
[MATH]B=\frac{1}{2}[/MATH]
[MATH]C=\frac{3}{2}[/MATH]
[MATH]D=\frac{5}{2}[/MATH]
[MATH]E=\frac{3}{2}[/MATH]
[MATH]F=0[/MATH]
[MATH]G=\frac{48}{7}[/MATH]
And so our particular solution is:
[MATH]p_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n[/MATH]
And so by the principle of superposition, we find:
[MATH]S_n=p_n+h_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n+c_1[/MATH]
Now we can use the initial sum to determine the parameter:
[MATH]S_1=\frac{1}{7}+\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\frac{3}{2}+\frac{48}{7}+c_1=13\implies c_1=0[/MATH]
And so we find the solution is:
[MATH]S_n=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{3}{2}n^5+\frac{5}{2}n^4+\frac{3}{2}n^3+\frac{48}{7}n[/MATH]
Or:
[MATH]S_n=\frac{n}{14}(2n^6+7n^5+21n^4+35n^3+21n^2+96)[/MATH]
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
It is nice to have conformation without the workings.