The Method of Differences
The Method of Differences
- If we have a series in the form of \(f(r)-f(r+1)\), then:
- \(\sum_{r=1}^{n}u_{r}=\sum_{r=1}^{n}(f(r)-f(r+1))\)
- \(u_{1} = f(1)-f(2)\)
- \(u_{2}=f(2)-f(3)\)
- \(\dots\)
- \(u_{n}=f(n)-f(n+1)\)
- This means that \(\sum_{r=1}^{n}=f(1)-f(n+1)\)
Example 1
- Show that \(4r^3\equiv r^2(r+1)^2-(r-1)^2r^2\)
- \(=r^2(r^2+2r+1)-(r^2-2r+1)r^2\)
- \(=r^4+2r^3+r^2-r^4+2r^3-r^2\)
- \(=4r^3\)
- Prove that \(\sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\)
- Consider\(\sum_{r=1}^{n}(r^2(r+1)^2-(r-1)^2r^2)\)
- Let \(r=1\): \(\cancel{ 1^2(2^2) }-0^2(1^2)\)
- Let \(r=2\): \(\cancel{ 2^2(3^2) }\cancel{ -1^2(2^2) }\)
- Let \(r=3\): \(3^2(4^2)\cancel{ -2^2(3^2) }\)
- \(\dots\)
- Let \(r=n\): \(n^2(n+1)^2\cancel{ -(n-1)^2n^2 }\)
- Leaves us with \(n^2(n+1)^2\)
- So: \(4\sum_{r=1}^{n}r^3=n^2(n+1)^2\)
- \(\therefore \sum_{r=1}^nr^3=\frac{1}{4}n^2(n+1)^2\)
1a) Show that \(r\equiv \frac{1}{2}(r(r+1))-r(r-1)\)
- \(r\equiv \frac{r^2+r}{2}-r^2+r\)
- \(2r\equiv r^2+r-r^2+r\)
- \(2r=2r\)
- \(r=r\)
1b) Hence show that \(\sum_{r=1}^{n}r=\frac{n}{2}(n+1)\) using the method of differences.
- \(\sum_{r=1}^{n}=\frac{1}{2}\sum_{r=1}^{n}r(r+1)-\frac{1}{2}\sum_{r=1}^{n}r(r-1)\)
- \(r=1\): \(\cancel{ \frac{1}{2}\times 1 \times 2 }-\frac{1}{2}\times 1 \times 0\)
- \(r=2\): \(\frac{1}{2}\times2\times3-\cancel{ \frac{1}{2}\times2\times 1 }\)
- \(\dots\)
- \(r=n\): \(\left( \frac{1}{2}\times n\times (n+1) \right)\cancel{ -(\frac{1}{2}\times n \times ((n-1)) })\)
- Leaves us with\(\frac{1}{2}n( n+1)\)