Question

If $1.(0!)+3.(1!)+7.(2!)+13.(3!)+21.(4!)+...$ upto $n$ terms$=\left(4000\right)4000!$ then the value of $n$ is

• A. $4000$
• B. $4001$
• C. $3999$
• D. $noneofthese$

Answer 1

Answer: (A)

$4000$

$1(0!)+3(1!)+7(2!)+13(3!)+.......nterms$

$t_n$ is given by

$t_n=(n^2-n+1)(n-1)!$

$=[n^2-(n-1\left)\right](n-1)!$

$=n^2(n-1)!-(n-1)(n-1)!$

$=n\left(n\right)!-(n-1)(n-1)!$

$G.E=S=\sum _{k=1}^{n}k\left(k\right)!-(k-1)(k-1)!$

$=1\left(1\right)!-0\left(0\right)!+2\left(2\right)!-1\left(1\right)!$

$+........+n\left(n\right)!-(n-1)(n-1)!$

$n\left(n\right)!$

But given that $S=n\left(n\right)!=4000\left(4000\right)!$

$\therefore n=4000$