Not hard

Algebra Level 3

1 1 × 2 + 1 2 × 3 + 1 3 × 4 + . . . . . . + 1 99 × 100 = a b \frac{1}{1 \times 2} +\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+......+\frac{1}{99 \times 100}=\frac{a}{b} where a b \frac{a}{b} is in simplest fractional form. Find a + b a+b . After finding a + b a+b , express it in the form of n 100 − x n100-x . What is n + x n+x ?

[Note: a , b , n , x ∈ Z + a,b,n,x\in\mathbb{Z^{+}} and 0 < n , x < 100 0<n,x<100 ]


The answer is 3.

Ritvik Vantipalli by Ritvik Vantipalli , 20, India

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2 solutions

1 1 × 2 + 1 2 × 3 + 1 3 × 4 + … + 1 99 × 100 \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times 4}+\ldots+\frac{1}{99\times100} = 1 1 − 1 2 + 1 2 − 1 3 + 1 3 − … − 1 99 + 1 99 − 1 100 = \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\ldots-\frac{1}{99}+\frac{1}{99}-\frac{1}{100} = 1 1 + ( 1 2 − 1 2 ) + ( 1 3 − 1 3 ) + … + ( 1 99 − 1 99 ) − 1 100 = \frac{1}{1}+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\ldots+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100} = 1 − 1 100 = 99 100 = 1-\frac{1}{100}=\frac{99}{100}

Take 99 + 100 = 199 99+100=199

199 = 100 n − x 199=100n-x ⇒ 100 n > 199 \Rightarrow 100n>199 ⇒ n = 2 \Rightarrow n=\boxed{2} 199 = 200 − x ⇒ x = 1 199=200-x \Rightarrow x=\boxed{1}

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@Ritzy Ritvik Why is it in Number Theory ?

Venkata Karthik Bandaru - 5 years, 9 months ago

I found the sum of n terms. For eg.- for the first term we have 1/2 then 2/3 then 3/4 .......so on. So there was a pattern of n/n+1 . The no. of terms are 99. So the answer is 199.

Satyajit Ghosh - 5 years, 9 months ago
Ramiel To-ong
May 12, 2016

nice analysis. same as mine.

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