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Question

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


Solution

$${T}_{n}$$ of $$1,4,7,...=1+\left(n-1\right)3=1+3n-3=3n-2$$ where $$a=1$$ and $$d=4-1=3$$
$${T}_{n}$$ of $$4,7,10...=4+\left(n-1\right)3=4+3n-3=3n+1$$ where $$a=3$$ and $$d=7-4=3$$
$${T}_{n}$$ of $$7,10,13...=7+\left(n-1\right)3=7+3n-3=3n+4$$ where $$a=7$$ and $$d=10-7=3$$
$$\therefore {T}_{n}=\dfrac{1}{6}\left[\dfrac{1}{\left(3n-2\right)\left(3n+1\right)\left(3n+4\right)}\right]$$
$${S}_{n}=\dfrac{1}{6}\dfrac{\left(3n+4\right)-\left(3n-2\right)}{\left(3n-2\right)\left(3n+1\right)\left(3n+4\right)}$$
$$=\dfrac{1}{6}\left[\dfrac{1}{\left(3n-2\right)\left(3n+1\right)}-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$$
We have $${S}_{n}=\sum{{T}_{r}}$$
                        $$=\dfrac{1}{6}\sum_{r=1}^{n}{\left[\dfrac{1}{\left(3n-2\right)\left(3n+1\right)}-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]}$$
                       $$=\dfrac{1}{6}\left[\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+....-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$$
 $$\therefore {S}_{n}=\dfrac{1}{6}\left[\dfrac{1}{4}-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$$
$$=\dfrac{1}{6}\left[\dfrac{\left(3n+1\right)\left(3n+4\right)-4}{4\left(3n+1\right)\left(3n+4\right)}\right]$$
$$=\dfrac{9{n}^{2}+12n+3n+4-4}{24\left(3n+1\right)\left(3n+4\right)}$$
$$=\dfrac{9{n}^{2}+15n}{24\left(3n+1\right)\left(3n+4\right)}$$
$$=\dfrac{3n\left(3n+5\right)}{24\left(3n+1\right)\left(3n+4\right)}$$
$$=\dfrac{n\left(3n+5\right)}{8\left(3n+1\right)\left(3n+4\right)}$$



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