# Calculate $(\frac12-\frac13)+(\frac13-\frac14)+...+(\frac19-\frac1{10})$

Calculate (\frac12-\frac13)+(\frac13-\frac14)+...+(\frac19-\frac1{10})

• 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’!

Show that \frac1{r^2}-\frac1{(r+1)^2}=\frac{2r+1}{r^2(r+1)^2}

• \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)

Hence solve \sum^n_{r=1} \frac{2r+1}{r^2 (r+1)^2}

• \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}.

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)
• =(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)