Question

How do you simplify $\dfrac{{(2k)!}}{{(2k + 2)!}}$ to $\dfrac{1}{{2(k + 1)(2k + 1)}}$?

Answer 1

Hint: The exclamation symbol in the question denotes the factorial symbol. It is a function where we find the product of all the constants below or equal to the given number. So, keep expanding the denominator till we get the numerator and then cancel the terms. Evaluate further till we get the expression we need to prove.

Complete step-by-step solution:
The given expression is, $\dfrac{{(2k)!}}{{(2k + 2)!}}$
Since factorial means writing the product of all the numbers equal to and below the given constant, we will now write the factorial of the expression in the denominator.
$\Rightarrow (2k + 2)! = (2k + 2)(2k + 1)(2k)!$
We will only expand it till we get the numerator term, $2\;k$ which later will be easier to simplify by canceling it out.
Now, write it all together and evaluate.
$\Rightarrow \dfrac{{(2k)!}}{{(2k + 2)!}} = \dfrac{{(2k)!}}{{(2k + 2)(2k + 1)(2k)!}}$
Now cancel the common terms from the numerator and the denominator.
The only term which is common in both the numerator and denominator is $(2k)!$
$\Rightarrow \dfrac{1}{{(2k + 2)(2k + 1)}}$
Now take $2$ commonly from $(2k + 2)$ to get our desired expression in the end.
$\Rightarrow \dfrac{1}{{2(k + 1)(2k + 1)}}$
$\therefore \dfrac{{(2k)!}}{{(2k + 2)!}} = \dfrac{1}{{(2k + 2)(2k + 1)}}$
LHS=RHS
Hence proved.

Additional Information: The factorial function is a mathematical function that is denoted by the exclamation mark $!\;$ . It is a function which when used finds out the product of all the numbers including the number given till the number $1$ . The factorial function is commonly used in permutations and combinations and even in finding probabilities.

Note: One should never forget to include the number self while writing the factorial of a given number. Whenever there is an expression where we must solve the factorials of given variables, just keep multiplying the product of the same variable but subtract $1$ from it every time you multiply.