Method of differences

You may notice that a lot of terms cancel out:
- \frac13 is a negative term in the first bracket, and a positive term in the second bracket.
- \frac14 is a negative term in the second bracket, and a positive term in the third bracket.
- This pattern continues for all the middle terms, the last being \frac19.
This leaves us with just \frac12 - \frac1{10}.
Simplifying this gives us \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac25.
Answer: \frac25.

This is called the ‘method of differences’!

\frac1{r^2}-\frac1{(r+1)^2} (write down the left-hand side)
\frac{(r+1)^2}{r^2 (r+1)^2} - \frac{r^2}{r^2 (r+1)^2} write each term with a common denominator
=\frac{(r+1)^2 - r^2}{r^2 (r+1)^2} (get a common denominator)
=\frac{(r^2 + 2r + 1) - r^2}{r^2 (r+1)^2} (expand the numerator)
=\frac{2r + 1}{r^2 (r+1)^2} (simplify)

\sum^n_{r=1} \frac{2r+1}{r^2 (r+1)^2}
=\sum^n_{r=1} \left( \frac1{r^2} - \frac1{(r+1)^2} \right) (use the result from above)
=\left(\frac1{1^2} - \frac1{2^2}\right) + \left(\frac1{2^2} - \frac1{3^2}\right) + \left(\frac1{3^2} - \frac1{4^2}\right) + ... +
\left(\frac1{(n-2)^2} - \frac1{(n-1)^2}\right) + \left(\frac1{(n-1)^2} - \frac1{n^2}\right) + \left(\frac1{n^2} - \frac1{(n+1)^2}\right)
(write out the first and last few terms)
All except the first and last terms cancel out, leaving us with \frac1{1^2} - \frac1{(n+1)^2}
=1 - \frac1{(n+1)^2} (simplify)
Answer: 1 - \frac1{(n+1)^2}.

\sum(24r^2+2)=\sum^n_{r=1}\left((2r+1)^3-(2r-1)^3\right)
=(3^2-1^2)+(5^2-3^2)+(7^2-5^2)+...+(2n+1)^2-(2n-1)^2+((2n+1)^2-(2n-1)^2) (write out the first few terms)
\sum(24r^2+2)=(2n+1)^3-1^3 (all middle terms cancel out)
24\sum r^2 + 2n = (2n+1)^3 - 1 (expand the left-hand side)
24\sum r^2 = (2n+1)^3 - 1 - 2n (rearrange)
24\sum r^2= 8n^3 + 12n^2 + 6n + 1 - 1 - 2n (expand the right-hand side)
24\sum r^2= 8n^3 + 12n^2 + 4n (simplify)
24\sum r^2=4n(2n^2 + 3n + 1) (factorise out a 4n)
24\sum r^2=4n(2n+1)(n+1) (factorise the quadratic)
\sum r^2 = \frac4{24}n(2n+1)(n+1) (divide both sides by 24)
\sum r^2=\frac16 n(n+1)(2n+1) (simplify)

flashcards

Question	Answer
Question	Answer
What is the method of differences?	A technique where terms in a sum cancel each other out, leaving only the first and last terms.
What is the value of (\frac12-\frac13)+(\frac13-\frac14)+...+(\frac19-\frac1{10})?	\frac25 (since all middle terms cancel to give \frac12 - \frac1{10} = \frac{4}{10} = \frac25).
Show that \frac1{r^2}-\frac1{(r+1)^2}=\frac{2r+1}{r^2(r+1)^2}.	\frac1{r^2}-\frac1{(r+1)^2} = \frac{(r+1)^2}{r^2 (r+1)^2} - \frac{r^2}{r^2 (r+1)^2} = \frac{(r+1)^2 - r^2}{r^2 (r+1)^2} = \frac{(r^2 + 2r + 1) - r^2}{r^2 (r+1)^2} = \frac{2r + 1}{r^2 (r+1)^2}
Hence solve \sum^n_{r=1} \frac{2r+1}{r^2 (r+1)^2}.	= \sum^n_{r=1} \left( \frac1{r^2} - \frac1{(r+1)^2} \right) (using the result), which telescopes to 1 - \frac1{(n+1)^2}.
Given that (2r+1)^3-(2r-1)^3=24r^2+2, show that \sum r^2=\frac16 n(n+1)(2n+1).	\sum(24r^2+2) = \sum^n_{r=1}\left((2r+1)^3-(2r-1)^3\right) = (2n+1)^3-1^3. So 24\sum r^2 + 2n = (2n+1)^3 - 1. Thus 24\sum r^2= (2n+1)^3 - 1 - 2n = 8n^3 + 12n^2 + 4n = 4n(2n^2 + 3n + 1) = 4n(2n+1)(n+1). Divide by 24: \sum r^2 = \frac16 n(n+1)(2n+1).
What remaining terms do you get after applying the method of differences to \sum^n_{r=1} (\frac1{r^2} - \frac1{(r+1)^2})?	\frac1{1^2} - \frac1{(n+1)^2}, i.e., 1 - \frac1{(n+1)^2}.