let the 3 terms be defined in the usual way of a a GS
a, ar, and ar^2
so
we have to show that
(a^2 + a^2 r^2 + a^2 r^4) / (a + ar + ar^2) divides evenly
that is, our denominator should cancel
(a^2 + a^2 r^2 + a^2 r^4) / (a + ar + ar^2)
= a^2(1 + r^2 + r^4)/( a(1+r + r^2) )
= a(1 + r^2 + r^4)/(1+r + r^2)
now 1+r^2 + r^4 is a GS , with a=1 , and common ratio of r^2
Sum(3) = 1( (r^2)^3 - 1)/(r^2 - 1)
= (r^6 - 1)/(r^2 - 1)
and 1 + r + r^2 is a GS, with a=1 and common ratio of r
sum(3) = (r^3-1)/(r-1)
so ....
a(1 + r^2 + r^4)/(1+r + r^2)
=a(r^6 - 1)/(r^2-1) รท ( (r^3 - 1/(r-1) )
= a(r^6 - 1)/(r^2-1) * (r-1)/(r^3-1)
= a(r^3-1)(r^3+1)(r-1)/( (r+1)(r-1)(r^3-1) )
= a(r^3 + 1)/(r+1)
= a(r+1)(r^2 - r + 1)/(r+1) ---- using sum of cubes factoring
= a(r^2 - r + 1) , no divisors left at the bottom
Hope somebody can come up with a shorter way.
CHECK:
eg. suppose we have 7, 28, 112
here a=7 , r = 4
a^2 + b^2 + c^2 = 7^2 + 28^2 + 112^2 = 13377
a+b+c = 7+28+112 = 147
13377/147 = 91
according to my formula the quotient should be
a(r^2 - r+1)
= 7(16-4+1) = 91 YEAHHHH!
If a, b, and c are three integers in geometric progression, prove that the number a^2+b^2+c^2 is exactly divisible by the number a+b+c.
2 answers
Looks good to me. That was going to be my approach until I saw it so ably demonstrated...
0 answers