Question #60895

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Answer 1 · 2016-10-31T00:20:55

#1007^2#

Calling #n=2014# we have

#2014³ - 2013³ + 2012³ - 2011³ + ... + 2³- 1³= sum_(k=1)^(n/2)(2k)^3-sum_(k=1)^(n/2)(2k-1)^3 #

but

#sum_(k=1)^n k^3 =((n(n+1))/2)^2#

Now

#sum_(k=1)^(n/2)(2k)^3=2^3sum_(k=1)^(n/2)k^3 = 2^3((n/2(n/2+1))/2)^2#

and

#sum_(k=1)^(n/2)(2k-1)^3=((n(n+1))/2)^2-2^3((n/2(n/2+1))/2)^2#

and

#sum_(k=1)^(n/2)(2k)^3-sum_(k=1)^(n/2)(2k-1)^3 =2xx2^3((n/2(n/2+1))/2)^2-((n(n+1))/2)^2#

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

in this case

#2n+3=4031=29 xx 139#

so the largest perfect square is #(2014/2)^2= 1007^2#