This can't be that Short!

Algebra Level 4

k=1m[(k2+1) k!]=1999×2000! \displaystyle \sum_{k=1}^m \left [ (k^2 + 1) \ k! \right ] = 1999 \times 2000!

What value of mm satisfies the above summation?

×

Problem Loading...

Note Loading...

Set Loading...