若等比数列前n项,前2n项,前3n项的和分别为sn s2n s3n 求证sn∧2+s2n∧2=sn(s2n+s3n)

an = a1q^(n-1)
Sn = a1(q^n-1)/(q-1)
(Sn)^2 + (S(2n))^2
= [a1(q^n-1)/(q-1)]^2 +[a1(q^(2n)-1)/(q-1)]^2
= [a1/(q-1)]^2 .[ q^(4n) -q^(2n) - 2q^n + 2]
Sn[S(2n) + S(3n) ]
=[a1(q^n-1)/(q-1)] .[ a1(q^(2n)-1)/(q-1) + a1(q^(3n)-1)/(q-1) ]
= [a1/(q-1)]^2 .(q^n - 1)[ q^(3n) +q^(2n) -2 ]
= [a1/(q-1)]^2 .[ q^(4n) -q^(2n) -2q^n +2 ]
=(Sn)^2 + (S(2n))^2