Basic calculation problem

Algebra Level 3

Evaluate:

1 x ( x + 1 ) + 1 ( x + 1 ) ( x + 2 ) + 1 ( x + 2 ) ( x + 3 ) + … + 1 ( x + 99 ) ( x + 100 ) + 1 x + 100 \frac{1}{x(x+1)} + \frac{1}{(x+1)(x+2)} + \frac{1}{(x+2)(x+3)}+\ldots+\frac{1}{(x+99)(x+100)} +\frac{1}{x+100}

1 x \frac{1}{x} 1 None of these choices 1 x − 1 100 \frac{1}{x}-\frac{1}{100}

Adam Phúc Nguyễn by Adam Phúc Nguyễn , 20, Vietnam

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2 solutions

1 x ( x + 1 ) + 1 ( x + 1 ) ( x + 2 ) + 1 ( x + 2 ) ( x + 3 ) + … + 1 ( x + 99 ) ( x + 100 ) + 1 x + 100 \frac{1}{x(x+1)} + \frac{1}{(x+1)(x+2)} + \frac{1}{(x+2)(x+3)}+\ldots+\frac{1}{(x+99)(x+100)} +\frac{1}{x+100}

= 1 x − 1 x + 1 + 1 x + 1 − 1 x + 2 + 1 x + 2 − 1 x + 3 + … − 1 x + 100 + 1 x + 100 =\frac{1}{x} - \frac{1}{x+1} + \frac{1}{x+1}-\frac{1}{x+2} + \frac{1}{x+2}-\frac{1}{x+3}+\ldots-\frac{1}{x+100}+\frac{1}{x+100}

Pair it on:

= 1 x + ( − 1 x + 1 + 1 x + 1 ) + ( − 1 x + 2 + 1 x + 2 ) + ( − 1 x + 3 + … + ( − 1 x + 100 + 1 x + 100 ) =\frac{1}{x}+( - \frac{1}{x+1} + \frac{1}{x+1})+(-\frac{1}{x+2} + \frac{1}{x+2})+(-\frac{1}{x+3}+\ldots+(-\frac{1}{x+100}+\frac{1}{x+100})

= 1 x = \boxed{\frac{1}{x}}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Did it the same way.Upvoted!

Athiyaman Nallathambi - 5 years, 9 months ago
Revanth Gumpu
Aug 24, 2015

Just rewrite each fraction as 1/n -1/(n+1) and then cancel out the same pairs and your left with 1/x.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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