Question

等比数列a4-a2=a2+a3=12求a1和q 求s6

Answer 1

a4-a2=a2(q²-1)=12≠0

q²≠1 q≠1且q≠-1

a4-a2=a2+a3

a2q²-a2=a2+a2q

等式两边同除以a2

q²-1=1+q

q²-q-2=0

(q-2)(q+1)=0

q=2或q=-1(舍去)

a4-a2=a2(q²-1)=3a2=12 a2=4 a1=a2/q=4/2=2

S6=a1(q^6 -1)/(q-1)

=2×(2^6 -1)/(2-1)

=2^7 -2

=128-2

=126