+0  
 
+13
364
9
avatar+-46 

How  should I start?

 Mar 15, 2019
 #5
avatar+109723 
0

\((1+x+x^2+......+x^{14})^2\\~\\ =1+2x+3x^2+ .........+15x^{14}+14x^{13}+13x^{12}+............2x^{27}+x^{28}\)

 

 

 

so the product is:

 

\((1+x+x^2+x^3+.......x^{27})(1+2x+3x^2+ .........+15x^{14}+14x^{13}+13x^{12}+............2x^{27}+x^{28})\\ \)

 

I am interested in the sum of the  x^28 coefficients.   

 

This is it

2+3+....+14+15  +14+13+......+2+1

=2(2+3+4+5+6+7+8+9+10+11+12+13+14)+15 +1

= 224

 

So the coefficient of  x^28=  224

 

-------------------

Maybe BaldisBasics knows a short cut but he certainly has not explained it to me. 

 Mar 15, 2019
 #6
avatar-46 
+9

That makes more sense, its a better explination. But thank you to both of you it really helped me out. 

LovinLife  Mar 15, 2019
 #7
avatar+109723 
-1

I expect there is a better way to do it.  :/

Melody  Mar 15, 2019
 #8
avatar+4582 
-1

BaldisBasics copied the solution from this website, and that is why some of the latex is not showing up.

https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19

 Mar 16, 2019
edited by tertre  Mar 16, 2019
 #9
avatar+109723 
-2

Thanks Tertre  That explains it. :)

Melody  Mar 16, 2019

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