Let's compute this manually!

Algebra Level 4

Let a a be a real number such that

a = 1 1 + 1 3 + 1 6 + 1 10 + ⋯ + 1 5050 . a=\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\cdots+\frac{1}{5050}.

If a a can be expressed in the form x y \frac{x}{y} , where x , y x,y are coprime, positive integers, find x + y x+y .


The answer is 301.

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Victor Loh by Victor Loh , 19, Singapore

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3 solutions

Yuxuan Seah
Jul 11, 2014

We have

1 1 + 1 3 + 1 6 + ⋯ + 1 5050 \frac{1}{1}+\frac{1}{3}+\frac{1}{6}+\cdots+\frac{1}{5050}

= 2 2 + 2 6 + 2 12 + ⋯ + 2 10100 =\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\cdots+\frac{2}{10100}

= 2 1 × 2 + 2 2 × 3 + 2 3 × 4 + ⋯ + 2 100 × 101 =\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+\cdots+\frac{2}{100\times101}

= 2 ( 1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 100 × 101 ) =2(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots+\frac{1}{100\times101})

= 2 ( 1 1 − 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + ⋯ + 1 100 − 1 101 ) =2(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{100}-\frac{1}{101})

= 2 ( 1 1 − 1 101 ) =2(\frac{1}{1}-\frac{1}{101})

= 2 ( 100 101 ) =2(\frac{100}{101})

= 200 101 =\boxed{\frac{200}{101}}

20 Helpful 0 Interesting 1 Brilliant 0 Confused

good method...

Mayank Holmes - 6 years, 10 months ago

did it using the same method

Kartik Sharma - 6 years, 10 months ago
Akshay Sant
Jul 10, 2014

Intrresting problem @Victor Loh

During summation of problem given I found an intresting pattern as follows: For "n"th term summation of series follows 2 n n + 1 \frac{2n}{n+1} . Thus for n=100,as sum of first 100 No.s is 5050 Thus by above statement Answer should be 200 101 \frac{200}{101} And thus 200+101=301 .... Thats It..

2 Helpful 0 Interesting 0 Brilliant 0 Confused

It should actually be 2 n ( n + 1 ) \dfrac{2}{n(n+1)} instead of 2 n n + 1 \frac{2n}{n+1} (I guess it's a typo).

This is no coincidence because it can easily be observed that the series is actually ∑ k = 1 100 1 T k \sum\limits_{k=1}^{100}\dfrac{1}{T_k} where T k T_k is the k th k^{\textrm{th}} triangular number . Using the well-known closed form for T k T_k and using telescopy, the problem becomes trivial.

Prasun Biswas - 5 years, 10 months ago
Michael Mendrin
Jul 10, 2014

Use induction for this one

2 x x + 1 + 1 1 2 ( x + 1 ) ( ( x + 1 ) + 1 ) = 2 ( x + 1 ) ( x + 1 ) + 1 \dfrac { 2x }{ x+1 } +\dfrac { 1 }{ \dfrac { 1 }{ 2 } (x+1)((x+1)+1) } =\dfrac { 2(x+1) }{ (x+1)+1 }

1 Helpful 1 Interesting 0 Brilliant 0 Confused

same actually i did it by same way @Michael Mendrin But during summation of problem given I found an intresting pattern as follows: For "n"th term summation of series follows 2 n n + 1 \frac{2n}{n+1} . Thus for n=100,as sum of first 100 No.s is 5050 Thus by above statement Answer should be 200 101 \frac{200}{101}

Akshay Sant - 6 years, 11 months ago

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