Q105 / 170 · Manipal (MET) 2021 · Mathematics

The value of $\frac{C(n, 2)}{(n+1)!}$ is

  • A
    $\frac{1}{2} e+1$
  • B
    $e+1$
  • Correct answer
    C
    $\frac{1}{2} e-1$
  • D
    $e$

Solution

Now, $\frac{{ }^n C_2}{(n+1)!}=\frac{\frac{n!}{(n-2)!2!}}{(\theta+1)!}$ $\begin{aligned} & =\frac{n!}{2(n+1)!(n-2)!}=\frac{n(n-1)}{2(n+1)!} \\ & =\frac{n^2-1+1-n}{2(n+1)!} \\ & =\frac{1}{2}\left[\frac{(n-7)}{n!}-\frac{1}{n!}+\frac{2}{(n+1)!}\right] \\ & =\frac{1}{2}\left[\frac{1}{(n-1)!}-\frac{2}{n!}+\frac{2}{(n+1)!}\right]\end{aligned}$ Put $n=0,1,2,3, \ldots$, we get $\begin{aligned} & =\frac{1}{2}[e-2(e)+2(e-1)] \\ & =\frac{1}{2}[e-2]=\frac{e}{2}-1\end{aligned}$

Hence, the correct answer is option: 'C'.

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