An algebra problem by Julian Yu

Algebra Level 5

How many terms are there when the expression ( x + y + z ) 2015 + ( x − y − z ) 2015 { (x+y+z) }^{ 2015 }+{ (x-y-z) }^{ 2015 } is expanded and simplified?


The answer is 1016064.

View wiki
Julian Yu by Julian Yu , 19, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rajen Kapur
Nov 17, 2015

Let the x a y b z c x^{a}y^{b}z^{c} be any term of the expression which being a sum of two expansions would have non-zero terms only in cases where (b + c) is even. As (a + b + c) = 2015, we can conclude that variable a can take values 1, 3, 5, ....., 2015 when (b + c) = (2015 - a) would be 2014, 2012, 2010, . . . . , 0. Corresponding to a given value of (b + c) say i there are (i +1) ways of assigning values (b, c) from (0, i), (1, i), (2, i), . . . ., (i, 1), (i, 0). Then the number of non-zero terms in the required expression are 1 + 3 + 5 + . . . . + 2015 = 100 8 2 1008^2 = 1016064.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you actually mean ( i + 1 ) (i +1) ways of assigning values ( b , c ) (b, c) from ( 0 , i ) , ( 1 , i − 1 ) , ( 2 , i − 2 ) , . . . . , ( i − 1 , 1 ) , ( i , 0 ) (0, i), (1, i-1), (2, i-2), . . . ., (i-1, 1), (i, 0) instead of the above. Other than that, great solution! :) I was stuck on this because I used the opposite yet more complex approach which is trying to find the complement (where a is even, then the terms cancel out). Yours is a much better and straightforward method!

Happy Melodies - 5 years, 6 months ago

nice solution .. :)

Ambuj Kumar Pandit - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...