Question

Sum the following series
$\frac{1}{\mathrm{1.4.7}}+\frac{1}{\mathrm{4.7.10}}+\frac{1}{\mathrm{7.10.13}}+....$ to $n$ terms

Answer 1

$T_n$ of $1,4,7,...=1+(n-1)3=1+3n-3=3n-2$ where $a=1$ and $d=4-1=3$

$T_n$ of $4,7,\mathrm{10...}=4+(n-1)3=4+3n-3=3n+1$ where $a=3$ and $d=7-4=3$

$T_n$ of $7,10,\mathrm{13...}=7+(n-1)3=7+3n-3=3n+4$ where $a=7$ and $d=10-7=3$

$\therefore T_n=\frac{1}{6}\left[\frac{1}{(3n-2)(3n+1)(3n+4)}\right]$

$S_n=\frac{1}{6}\frac{(3n+4)-(3n-2)}{(3n-2)(3n+1)(3n+4)}$

$=\frac{1}{6}[\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}]$

We have $S_n=\sum T_r$

$=\frac{1}{6}{\sum }_{r=1}^n[\frac{1}{(3n-2)(3n+1)}-\frac{1}{(3n+1)(3n+4)}]$

$=\frac{1}{6}[\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+....-\frac{1}{(3n+1)(3n+4)}]$

$\therefore S_n=\frac{1}{6}[\frac{1}{4}-\frac{1}{(3n+1)(3n+4)}]$

$=\frac{1}{6}\left[\frac{(3n+1)(3n+4)-4}{4(3n+1)(3n+4)}\right]$

$=\frac{9n^2+12n+3n+4-4}{24(3n+1)(3n+4)}$

$=\frac{9n^2+15n}{24(3n+1)(3n+4)}$

$=\frac{3n(3n+5)}{24(3n+1)(3n+4)}$

$=\frac{n(3n+5)}{8(3n+1)(3n+4)}$