Question Number 57251 by ANTARES VY last updated on 01/Apr/19
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19
![T_n =((n(n−1))/((1/((n−1)!))+(1/((n−2)!)))) =((n(n−1))/((1+n−1)/((n−1)!)))→(n−1)(n−1)! [Σ_(n=3) ^(2019) (n−1)(n−1)!]+2→S+2 T_3 =(2)×2! T_4 =(3)×3! T_5 =(4)×4! ... ... T_(2019) =(2018)×(2018)! so required sum total= 2+Σ_(n=3) ^(2019) (n−1)(n−1)! others pls check...](https://www.tinkutara.com/question/Q57256.png)
$${T}_{{n}} =\frac{{n}\left({n}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}+\frac{\mathrm{1}}{\left({n}−\mathrm{2}\right)!}} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)}{\frac{\mathrm{1}+{n}−\mathrm{1}}{\left({n}−\mathrm{1}\right)!}}\rightarrow\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}\right)! \\ $$$$\left[\underset{{n}=\mathrm{3}} {\overset{\mathrm{2019}} {\sum}}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}\right)!\right]+\mathrm{2}\rightarrow\boldsymbol{{S}}+\mathrm{2} \\ $$$$ \\ $$$${T}_{\mathrm{3}} =\left(\mathrm{2}\right)×\mathrm{2}! \\ $$$${T}_{\mathrm{4}} =\left(\mathrm{3}\right)×\mathrm{3}! \\ $$$${T}_{\mathrm{5}} =\left(\mathrm{4}\right)×\mathrm{4}! \\ $$$$… \\ $$$$… \\ $$$${T}_{\mathrm{2019}} =\left(\mathrm{2018}\right)×\left(\mathrm{2018}\right)! \\ $$$${so}\:{required}\:{sum}\:{total}= \\ $$$$\mathrm{2}+\underset{{n}=\mathrm{3}} {\overset{\mathrm{2019}} {\sum}}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}\right)! \\ $$$${others}\:{pls}\:{check}… \\ $$