
How do you simplify $ \dfrac{{(2n + 3)!}}{{(2n)!}} $ ?
Answer
560.7k+ views
Hint: In this question, we have to find the ratio of factorial of (2n+3) and factorial of 2n. To simplify this expression, we first have to understand what factorial actually is. The factorial is represented as $ a! $ where a is any positive integer, it indicates the product of all the positive integers smaller than or equal to a. We will expand the factorial in the numerator and then cancel out the common terms of the numerator and the denominator.
Complete step-by-step answer:
We have to simplify $ \dfrac{{(2n + 3)!}}{{(2n)!}} $
$ (2n + 3)! = (2n + 3)(2n + 2)(2n + 1)(2n)! $
Put this value in the given fraction, we get –
$ \dfrac{{(2n + 3)!}}{{(2n)!}} = \dfrac{{(2n + 3)(2n + 2)(2n + 1)(2n)!}}{{(2n)!}} $
Now, we see that $ (2n)! $ is present in both the numerator and the denominator of this fraction so we divide both the numerator and denominator by $ (2n)! $ and get –
$ \Rightarrow \dfrac{{(2n + 3)!}}{{(2n)!}} = (2n + 3)(2n + 2)(2n + 1) $
Now,
$
(2n + 3)(2n + 2)(2n + 1) = 8{n^3} + 24{n^2} + 22n + 6 \\
\Rightarrow \dfrac{{(2n + 3)!}}{{(2n)!}} = 8{n^3} + 24{n^2} + 22n + 6 \;
$
Hence the simplified form of $ \dfrac{{(2n + 3)!}}{{(2n)!}} $ is $ 8{n^3} + 24{n^2} + 22n + 6 $ .
So, the correct answer is “ $ 8{n^3} + 24{n^2} + 22n + 6 $ ”.
Note: Factorial is an important term to solve the questions related to permutation and combination. They convert many complex equations and activities simpler. After simplifying the fraction, we had to find the product of three parenthesis containing different terms, for that we first solve the first two brackets. We multiply each term of the first bracket with the second bracket one by one and then similarly we multiply each term of the obtained terms with the third bracket.
The product of the first and second bracket in the given solution is – $ (2n + 3)(2n + 2) = 2n(2n + 2) + 3(2n + 2) = 4{n^2} + 10n + 6 $
Then we multiply the obtained answer with the third bracket –
$ (4{n^2} + 10n + 6)(2n + 1) = 8{n^3} + 24{n^2} + 22n + 6 $
The expression cannot be simplified further, so this is the correct answer.
Complete step-by-step answer:
We have to simplify $ \dfrac{{(2n + 3)!}}{{(2n)!}} $
$ (2n + 3)! = (2n + 3)(2n + 2)(2n + 1)(2n)! $
Put this value in the given fraction, we get –
$ \dfrac{{(2n + 3)!}}{{(2n)!}} = \dfrac{{(2n + 3)(2n + 2)(2n + 1)(2n)!}}{{(2n)!}} $
Now, we see that $ (2n)! $ is present in both the numerator and the denominator of this fraction so we divide both the numerator and denominator by $ (2n)! $ and get –
$ \Rightarrow \dfrac{{(2n + 3)!}}{{(2n)!}} = (2n + 3)(2n + 2)(2n + 1) $
Now,
$
(2n + 3)(2n + 2)(2n + 1) = 8{n^3} + 24{n^2} + 22n + 6 \\
\Rightarrow \dfrac{{(2n + 3)!}}{{(2n)!}} = 8{n^3} + 24{n^2} + 22n + 6 \;
$
Hence the simplified form of $ \dfrac{{(2n + 3)!}}{{(2n)!}} $ is $ 8{n^3} + 24{n^2} + 22n + 6 $ .
So, the correct answer is “ $ 8{n^3} + 24{n^2} + 22n + 6 $ ”.
Note: Factorial is an important term to solve the questions related to permutation and combination. They convert many complex equations and activities simpler. After simplifying the fraction, we had to find the product of three parenthesis containing different terms, for that we first solve the first two brackets. We multiply each term of the first bracket with the second bracket one by one and then similarly we multiply each term of the obtained terms with the third bracket.
The product of the first and second bracket in the given solution is – $ (2n + 3)(2n + 2) = 2n(2n + 2) + 3(2n + 2) = 4{n^2} + 10n + 6 $
Then we multiply the obtained answer with the third bracket –
$ (4{n^2} + 10n + 6)(2n + 1) = 8{n^3} + 24{n^2} + 22n + 6 $
The expression cannot be simplified further, so this is the correct answer.
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