Q172 / 270 · NDA 2021 (Phase 2) · Mathematics

If $a_n=n(n!)$, then what is $a_1+a_2+a_3+\ldots+a_{10}$ equal to?

  • A
    $10!-1$
  • B
    $11!+1$
  • C
    $10!+1$
  • Correct answer
    D
    $11!-1$

Solution

$\begin{aligned} & \therefore a_n=n(n!)=(n+1) n!-n! \\ & =(n+1)!-n! \end{aligned}$ Now $a_1=2$ ! -1 ! $\begin{aligned} & a_2=3!-2! \\ & a_3=4!-3! \end{aligned}$ $a_{10} \overline{=11!-\ldots 10!}$ Adding all above equations we get $a_1+a_2+a_3 \ldots .+a_{10}=11!-1!$

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

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