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Question

Show that
$$\dfrac { 3\cdot 3 }{ 1+ } \dfrac { 3\cdot 4 }{ 2+ } \dfrac { 3\cdot 5 }{ 3+ } \cdots \dfrac { 3\left( n+2 \right)  }{ n+ } \cdots =\dfrac { 6\left( 2{ e }^{ 3 }+1 \right)  }{ 5{ e }^{ 3 }-2 } $$.


Solution

$${ u }_{ n }=n{ u }_{ n-1 }+3\left( n+2 \right) { u }_{ n-2 };$$
$${ u }_{ n }-\left( n+3 \right) { u }_{ n-1 }=-3\left\{ { u }_{ n-1 }-\left( n+2 \right) { u }_{ n-2 } \right\} ;$$
...........................................
$${ u }_{ 3 }-6{ u }_{ 2 }=-3\left( { u }_{ 2 }-5{ u }_{ 2 } \right) $$
By multiplication;
$${ u }_{ n }-\left( n+3 \right) { u }_{ n-1 }={ \left( -3 \right)  }^{ n-2 }\left( { u }_{ 2 }-5{ u }_{ 1 } \right) $$
Now,
$${ p }_{ 1 }=9,{ q }_{ 1 }=1,{ p }_{ 2 }=18,{ q }_{ 2 }=14,$$
$${ p }_{ n }-\left( n+3 \right) { p }_{ n-1 }={ \left( -1 \right)  }^{ n-1 }{ 3 }^{ n+1 }$$
$$\cfrac { { p }_{ n } }{ \left( n+3 \right) ! } -\cfrac { { p }_{ n-1 } }{ \left( n+2 \right) ! } =\cfrac { { \left( -1 \right)  }^{ n-1 }{ 3 }^{ n+1 } }{ \left( n+3 \right) ! } $$
.....................
$$\cfrac { { p }_{ 2 } }{ 5! } -\cfrac { { p }_{ 1 } }{ 4! } =\cfrac { \left( -1 \right) { 3 }^{ 2 } }{ 5! } $$
By addition;
$$\cfrac { { p }_{ n } }{ \left( n+3 \right) ! } =\cfrac { { 3 }^{ 2 } }{ 4! } -\cfrac { { 3 }^{ 3 } }{ 5! } +\cfrac { { 3 }^{ 4 } }{ 6! } -.....$$
$$=\cfrac { 1 }{ 9 } \left( \cfrac { { 3 }^{ 4 } }{ 4! } -\cfrac { { 3 }^{ 5 } }{ 5! } +\cfrac { { 3 }^{ 6 } }{ 6! } -..... \right) ;$$
and $$\cfrac { { q }_{ n } }{ \left( n+3 \right) ! } -\cfrac { 1 }{ 4! } =\cfrac { { 3 }^{ 2 } }{ 5! } -\cfrac { { 3 }^{ 3 } }{ 6! } +\cfrac { { 3 }^{ 4 } }{ 7! } -.....$$
$$=\cfrac { 1 }{ 27 } \left( \cfrac { { 3 }^{ 5 } }{ 5! } -\cfrac { { 3 }^{ 6 } }{ 6! } +\cfrac { { 3 }^{ 7 } }{ 7! } -..... \right) $$
$${ e }^{ -3 }=1-3+\cfrac { { 3 }^{ 2 } }{ 2! } -\cfrac { { 3 }^{ 3 } }{ 3! } +......$$
$$=-2+\cfrac { { 3 }^{ 4 } }{ 4! } -\cfrac { { 3 }^{ 5 } }{ 5! } +\cfrac { { 3 }^{ 6 } }{ 6! } -......$$
$$\therefore \cfrac { { p }_{ n } }{ { q }_{ n } } =\cfrac { 1 }{ 9 } \left( { e }^{ -3 }+2 \right) \div \cfrac { 1 }{ 27 } \left( \cfrac { 27 }{ 4! } -2+\cfrac { { 3 }^{ 4 } }{ 4! } -{ e }^{ -3 } \right) $$
$$\cfrac { \left( { e }^{ -3 }+2 \right)  }{ -2{ e }^{ -3 } } =\cfrac { 6\left( 2{ e }^{ 3 }+1 \right)  }{ 5{ e }^{ 3 }-2 } $$

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