(4+1)!=(4+1).5!

asked by guest
on Jan 27, 2025 at 12:55 am



You asked:

Investigate the equation: \(\left( 4 + 1 \right)! = \left(4 + 1\right) \cdot .5!\).

MathBot Answer:

This equation has no unknown quantities, so it is either true or false.

This equation is false because the left-hand side is not equal to the right-hand side.

The left-hand side of the equation is:

\[\left( 4 + 1 \right)! = 120\]

The right-hand side of the equation is:

\[\left(4 + 1\right) \cdot .5! = 5 \left(\frac{1}{2}\right)!\]