Alg9.3.2

From Exampleproblems

Show that $\frac{1\cdot2}{1!}+\frac{2\cdot3}{2!}+\frac{3\cdot4}{3!}+...........=3e\,$

nth term in the given series is $T_n=\frac{n(n+1)}{n!}\,$

Simplifying further, $T_n=\frac{n+1}{(n-1)!}\,$

$T_n=\frac{(n-1)+2}{(n-1)!}\,$

Splitting the fraction into two,we get $T_n=\frac{(n-1)}{(n-1)!}+\frac{2}{(n-1)!}\,$

Giving the values 1,2,3,4..... to n and adding all of them we get $T_n=e+2e=3e\,$