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Senior two must have four math problem series.
A:

1.

A 1=3, {an} is arithmetic progression, so a1+a2+…+an = (3+2n+1) n/2 = n (n+2).

So bn=n(n+2)/n=n+2, b 1=3.

Therefore, the sum of n terms before bn is Sn=(3+n+2)n/2=n(n+5)/2.

2.

When n= 1, s1= a1= 3a1/2-3, the solution is a 1=6.

When n≥2, an=Sn-S(n- 1)

=3/2an-3-(3/2a(n- 1)-3)

=3/2an-(3/2a(n- 1)

So 3a(n- 1)=an.

So {an} is a geometric series with the first term of 6 and the common ratio of 3.

an=6×3^(n- 1)=2×3^n

So an = 2× 3 n