Simplify (2x + 4x + 6x +8x +... +2006x + 2008x) x +3x 5x + 7x +...+ 2005x + 2007x).?

Question

2096 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T18:30:43

#(2x+4x+...+2006x+2008x)(x+3x+...+2005x+2007x)=1004^3(1004+1)x^2#

Firstly, take the common factor #x# to the front:

#(2x+...+2008x)(x+...+2007x)=x^2(2+4+...+2006+2008)(1+3+...+2005+2007)#

Let's denote the two sums as #S_1# and #S_2#.

#S_1 = 2+4+...+2006+2008#

#= 2(1+2+...+1003+1004)#

Remember the formula:

#1+2+3+...+(n-2)+(n-1)+n = (n(n+1))/2#

#:. S_1 = cancel2((color(red)(1004)(color(red)(1004)+1))/cancel2)=1004*1005=1009020#

The second sum is a bit trickier:

#S_2 = 1+3+...+2005+2007#

To transform this into a more familiar sum, we have to add a few terms:

#S_3= 1 + color(red)2+3+color(red)4+...+2005+color(red)2006+2007+color(red)2008#

Notice how the terms in red sum up to #S_1#, hence

#S_3 = S_2 + S_1 => S_2 = S_3 - S_1#

#S_3 = (color(red)(2008)(color(red)(2008)+1))/2=2017036#

#S_2 = 2017036 - 1009020 = 1008016#

#:. (2x+...+2008x)(x+...+2007x)#

#=x^2(S_1 * S_2)=(1009020*1008016)x^2#

Which is a pretty large number.

We can generalize this:

#{2x+4x+...+2(n-1)x+2nx} xx {x+3x+...++(2n-3)x +(2n-1)x}#

#=n^3(n+1)x^2#

With this, we can also realize that

#1009020*1008016=1004^3(1004+1)#