Daca a,b,c,d sunt in progresie aritmetica, atunci demonstrati ca:

a la puterea 3 + d la puterea 3 = b la puterea 3 + c la puterea 3 + 3(b-c)(ab-cd)
Ajutati-ma va rog frumos!

b=a+r, c=a+2r, d=a+3r
a^3+(a+3r)^3=(a+r)^3+(a+2r)^3+ 3(a+r-a-2r)[a(a+r)-(a+2r)(a+3r)]
a^3+a^3+3a^2*3r+3a*3^2*r^2+27r^3=a^3+3a^2*r+3a*r^2+r^3+a^3+3a^2*2r+3a*4*r^2+8r^3-3r[a^2+ar-(a^2+3ar+2ar+6r^2)]=
scad 2a^3 in ambele parti
9ra^2+27ar^2+27r^3=3ra^2+3ar^2+r^3+6ra^2+12ar^2+8r^3-3ra^2-3ar^2+3ra^2+9ar^2+6ar^2+18r^3
scad 27 r^3 in ambele parti si reducem termenii asemenea
9ra^2+27ar^2=9ra^2+27ar^2
am demonstrat ca este adevarat

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Albatran Albatran · Answer 2 · 2016-12-02T00:38:32+02:00

daca a,b,c,d in progresie aritmetica, atunci c-a=d-b=2r

c-a=d-b ridicam la puterea a treia

c³-3c²a+3ca²-a³=d³-3d²b+3db²-b³
c³-a³ +3ca(a-c)=d³-b³+3db(b-d)

c³+b³+3ac(a-c)-3bd(b-d)=d³+a³

d³+a³=c³+b³+3ac(a-c)-3bd(b-d)

observam ca am obtinut termenii din egalitatea ceruta doar daca
3(b-c)(ab-cd)=3ac(a-c)-3bd(b-d)
adica , impatind prin 3
(b-c)(ab-cd)=ac(a-c) +bd(d-b)
dar b-c=-r
a-c=-2r
d-b=2r
deci
-r(ab-cd)=ac* (-2r) +bd*(2r) impartim ecuatia cu -r
ab-cd=2ac-2bd
cd-ab=2(bd-ac)
(a+2r) (a+3r)- a(a+r)= 2 [(a+r)(a+3r)-a(a+2r)]
a²+5ar+6r²-a²-ar= 2(a²+4ar+3r²-a²-2ar)
4ar+6r²=2(2ar+3r²)
4ar+6r²=4ar+6r²
Adevarat, deci egalitatea ceruta este adevarata