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Which of the following is equal to x!/(x+1)! + (x+1)!/(x+2)!

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Which of the following is equal to x!/(x+1)! + (x+1)!/(x+2)!

by Max@Math Revolution » Thu Apr 12, 2018 12:48 am
[GMAT math practice question]

Which of the following is equal to
$$\frac{x!}{\left(x+1\right)!}+\frac{\left(x+1\right)!}{\left(x+2\right)!}?$$

$$A.\ 1+\frac{1}{x}$$
$$B.\ 1+\frac{1}{\left(x+1\right)}$$
$$C.\ \frac{\left(2x+3\right)}{\left(x+1\right)\left(x+2\right)}$$
$$D.\ \frac{1}{x\left(x+1\right)}$$
$$E.\ \frac{\left(2x-1\right)}{\left(x+1\right)\left(x+2\right)}$$
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by GMATGuruNY » Thu Apr 12, 2018 2:41 am
Max@Math Revolution wrote:[GMAT math practice question]

Which of the following is equal to
$$\frac{x!}{\left(x+1\right)!}+\frac{\left(x+1\right)!}{\left(x+2\right)!}?$$

$$A.\ 1+\frac{1}{x}$$
$$B.\ 1+\frac{1}{\left(x+1\right)}$$
$$C.\ \frac{\left(2x+3\right)}{\left(x+1\right)\left(x+2\right)}$$
$$D.\ \frac{1}{x\left(x+1\right)}$$
$$E.\ \frac{\left(2x-1\right)}{\left(x+1\right)\left(x+2\right)}$$
Let x=1.
Plugging x=1 into the given expression, we get:
$$\frac{1!}{\left(1+1\right)!}+\frac{\left(1+1\right)!}{\left(1+2\right)!}= {\frac{1}{2}}{ } + {\frac{1}{3}}{ } = {\frac{5}{6}}{ } $$
The target value is 5/6.
Now plug x=1 into the answer choices to see which yields the target value of 5/6.
Only C works:
$$\ \frac{\left(2*1+3\right)}{\left(1+1\right)\left(1+2\right)} = {\frac{5}{6}}{ } $$

The correct answer is C.
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by Max@Math Revolution » Sun Apr 15, 2018 5:32 pm
=>
x! / ( x + 1 )! + ( x + 1 )! / ( x + 2 )!
= x! / {x! * (x+1)} + (x+1)! / { (x+1)! * (x+2) }
= 1/(x+1) + 1/(x+2) = (2x+3)/(x+1)(x+2)

Therefore, C is the answer.
Answer: C
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