Question

Find the sum of n terms of the series $\frac{1}{\lfloor 3}+\frac{5}{\lfloor 4}+\frac{11}{\lfloor 5}+\frac{19}{\lfloor 6}+....$

Answer 1

$\Rightarrow t_n=\frac{n^2+n-1}{(n+2)!}$

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

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

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

$t_{n-1}=\frac{n-1}{n!}-\frac{n}{(n+1)!}$

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$t_1=-\frac{1}{2!}-\frac{2}{3!}$$S_n=\frac{1}{2}-\frac{n+1}{(n+2)!}$