Question

Find the sum of the following series $\frac{2}{\lfloor 1}+\frac{3}{\lfloor 2}+\frac{6}{\lfloor 3}+\frac{11}{\lfloor 4}+\frac{18}{\lfloor 5}+....$

Answer 1

The general term of $2,3,6,11,18,...$ can be obtained by order of differences and it is found to be $n^2-2n+3$

$\therefore u_n=\frac{n^2-2n+3}{n!}$

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

$u_1=\frac{3}{1}-1$

$u_2=\frac{3}{2!}-1+1$

$u_3=\frac{3}{3!}-\frac{1}{2!}+\frac{1}{1!}$

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....$

$e^1=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....$

$\therefore S=3(e-1)-e+e$

$=3e-3$