Converge?

Algebra Level 2

Let S \Large{S} = 1 1.2 + 1 2.3 + . . . . . . . . . . + 1 99.100 \dfrac{1}{1.2} + \dfrac{1}{2.3} + .......... + \dfrac{1}{99.100}

S \Large{S} can be expressed as a b \frac{a}{b} , where a , b a,b are co-prime integers.

Find b − a b-a .

NOTE -: a . b = a × b a.b = a \times b


The answer is 1.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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2 solutions

Sharky Kesa
Oct 7, 2014

Notice how you can rewrite the series as:

S = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) + . . . ( 1 99 − 1 100 ) S = (\dfrac {1}{1} - \dfrac {1}{2} ) + (\dfrac {1}{2} - \dfrac {1}{3}) + ... ( \dfrac {1}{99} - \dfrac {1}{100} )

Which can be simplified as:

S = 1 1 − 1 2 + 1 2 − 1 3 + . . . 1 99 − 1 100 S = \dfrac {1}{1} - \dfrac {1}{2} + \dfrac {1}{2} - \dfrac {1}{3} + ... \dfrac {1}{99} - \dfrac {1}{100}

S = 1 1 − 1 100 S = \dfrac {1}{1} - \dfrac {1}{100}

S = 99 100 S = \dfrac {99}{100}

100 − 99 = 1 100 - 99 = 1

Therefore, the answer is 1.

11 Helpful 0 Interesting 0 Brilliant 0 Confused

A nice solution @Sharky Kesa ,solved the same way.

Harsh Shrivastava - 6 years, 8 months ago

@Sharky Kesa you and I just posted two diff solutions exactly at the same time .

Shubhendra Singh - 6 years, 8 months ago

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Yeah, I noticed. :D

Sharky Kesa - 6 years, 8 months ago

Is it telescopic sum?

Aman Real - 6 years, 1 month ago

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Yes, it's a telescopic sum.

Harsh Shrivastava - 6 years, 1 month ago
André Luiz
Jun 21, 2015

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