Question

If ${\sum }_{k=1}^m(k^2+1)(k!)=1999(2000!)$, then the value of $m$ is

• A. $1999$
• B. $2000$
• C. $2001$
• D. $2002$

Answer 1

Answer: (A)

$1999$

${(k}^2+1\left)\right(k!)={\left(\right(k}^2+k)-(k-1)\left)\right(k!)$
$=k(k+1)!-(k-1)k!$
This is the form required to form a regression. Notice how this one term of the sequence is 'made up of' two parts - and one part can be made by making a simple substitution into the other part (for example, $k$ to $k+1$).
Now, when the next term in the sequence is added, it cancels out one part of the previous, and leaves only two 'parts':
$\left[k\right(k+1)!-(k-1)k!]+$
$\left[\right(k+1\left)\right(k+2)!-k(k+1)!]$
$=(k+1)(k+2)!-(k-1)(k!)$
This goes on continuously, from the first to the last term, and in the end, the surviving terms are:
$-(1-1)(1!)+m(m+1)!=m(m+1)!$
So, ${\sum }_{k=1}^m(k^2+1)(k!)=m(m+1)!=1999(2000!)$
$\Rightarrow m=1999$