  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 $${S}_{n}=\sum{{T}_{r}}$$                        $$=\dfrac{1}{6}\sum_{r=1}^{n}{\left[\dfrac{1}{\left(3n-2\right)\left(3n+1\right)}-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]}$$                       $$=\dfrac{1}{6}\left[\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+....-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$$ $$\therefore {S}_{n}=\dfrac{1}{6}\left[\dfrac{1}{4}-\dfrac{1}{\left(3n+1\right)\left(3n+4\right)}\right]$$$$=\dfrac{1}{6}\left[\dfrac{\left(3n+1\right)\left(3n+4\right)-4}{4\left(3n+1\right)\left(3n+4\right)}\right]$$$$=\dfrac{9{n}^{2}+12n+3n+4-4}{24\left(3n+1\right)\left(3n+4\right)}$$$$=\dfrac{9{n}^{2}+15n}{24\left(3n+1\right)\left(3n+4\right)}$$$$=\dfrac{3n\left(3n+5\right)}{24\left(3n+1\right)\left(3n+4\right)}$$$$=\dfrac{n\left(3n+5\right)}{8\left(3n+1\right)\left(3n+4\right)}$$Mathematics

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 