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Algebra Level 3

If the value of

( 1 + 1 2 + 1 3 + ⋯ + 1 2014 ) ( 1 2 + 1 3 + 1 4 + ⋯ + 1 2015 ) \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2014}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2015}\right) − ( 1 + 1 2 + 1 3 + ⋯ + 1 2015 ) ( 1 2 + 1 3 + 1 4 + ⋯ + 1 2014 ) -\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2014}\right)

is a b \frac{a}{b} for relatively prime a a and b b , then the value of a + b a+b is

2015 2013 2014 2016

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Tom Zhou by Tom Zhou , 21, Canada

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5 solutions

Let x = 1 + 1 2 + 1 3 + . . . + 1 2014 x = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2014} and y = x + 1 2015 y = x + \frac{1}{2015} .

Then the expression becomes

x ∗ ( y − 1 ) − y ∗ ( x − 1 ) = y − x = 1 2015 x*(y - 1) - y*(x - 1) = y - x = \frac{1}{2015} .

Thus a = 1 , b = 2015 a = 1, b = 2015 and a + b = 2016 a + b = \boxed{2016} .

32 Helpful 0 Interesting 0 Brilliant 0 Confused

nice solution, very clear thanks

Mardokay Mosazghi - 6 years, 9 months ago

same , but i let y= 1 / 2 + 1 / 3 + 1 / 4 + . . . . . . . . + 1 / 2015 1/2+1/3+1/4+........+1/2015

math man - 6 years, 9 months ago

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i too did same

Rishabh Jain - 6 years, 9 months ago

why u are substituting 1 from y,x ? means, why (y-1),(x-1)

Roskstar Rudra - 6 years, 9 months ago

awsome solution

Saptakatha Adak - 6 years, 8 months ago

great one :)

Eve Gayle Garcia - 6 years, 8 months ago

Well done, crisp and clear solution!

sandeep reddy - 6 years, 7 months ago
Jack Mamati
Oct 14, 2014

Here's my solution.

Let x = 1 2 + 1 3 + 1 4 + . . . + 1 2014 x=\frac {1}{2} +\frac{1}{3} +\frac{1}{4}+...+ \frac{1}{2014}

The expression will become

( 1 + x ) ( x + 1 2015 ) − ( 1 + x + 1 2015 ) ( x ) (1+x)(x+\frac{1}{2015})-(1+x+\frac{1}{2015} )(x)

= x 2 + x + x 2015 + 1 2015 − x − x 2 − x 2015 =x^2+x+\frac{x}{2015}+\frac{1}{2015}-x-x^2-\frac{x}{2015}

= 1 2015 = a b =\frac{1}{2015}=\frac{a}{b}

a + b = 2016 a+b=2016 (answer)

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Sriharsha Dv
Sep 18, 2014

But 2015 is not a prime number(divisible by 5)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

If two numbers are " relatively prime ", then there is no positive integers other than 1 that can divide both numbers

Kenny Lau - 6 years, 8 months ago
Rohit Sachdeva
Sep 12, 2014

I too did similarly..

1 2 + 1 3 + . . . 1 2014 = x \frac{1}{2}+\frac{1}{3}+...\frac{1}{2014}=x

a n d 1 2015 = y and \frac{1}{2015}=y

Then expression becomes:

( 1 + x ) ( x + y ) − x ( 1 + x + y ) (1+x)(x+y)-x(1+x+y)

= y =y

So a=1 & b=2015; a+b=2016

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This is my solution:

Let x = 1/2 +1/3 + 1/4 +...+1/2014

then the equation becomes,

(1 + x)(x + 1/2015) - (2016/2015 + x)(x) = a/b

x^2 + x + x/2015 + 1/2015 - (2016x/2015 + x^2) = a/b

x^2 + 2016x/2015 + 1/2015 - 2016x/2015 - x^2 = a/b

1/2015 = a/b

therefore, a = 1, b = 2015...

a + b = 1 + 2015 = 2016

Mark Angelo Sibayan - 6 years, 8 months ago
William Isoroku
Nov 20, 2014

Substituting a=(1/2+1/3+....... 1/2015) and b=(1/2+1/2+.......1/2014) the expression becomes (1+b)a-(1+a)b which simplifies to a-b. Since a=b+1/2015, a-b=1/2015. 1+2015=2016.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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