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Question

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


Solution

$${ n }^{ th }$$ term $$T_{n}=\cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) \left( 3n+4 \right)  } $$
By partial fraction we get
$${ T }_{ n }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right)  } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right)  }  \right] $$
$${ T }_{ 1 }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 1.4 } -\cfrac { 1 }{ 4.7 }  \right] $$
$${ T }_{ 2 }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 4.7 } -\cfrac { 1 }{ 7.10 }  \right] $$
$$ { T }_{ n }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right)  } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right)  }  \right]$$
Then, $${ S }_{ n }=\displaystyle \sum_{n=1}^{n} T_{n}=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 1.4 } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right)  }  \right] $$
$$=\cfrac { 1 }{ 24 } -\cfrac { 1 }{ 6\left( 3n+1 \right) \left( 3n+4 \right)  } $$
As $$n\rightarrow \infty $$
$$S_{n}=\cfrac { 1 }{ 24 } -0$$
$$=\cfrac { 1 }{ 24 } $$

Mathematics

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