数列{an},{bn}中,a1=3,b1=0,当n≥2时an=[2a(n-1)+b(n-1)]/3,bn=[a(n-1)+2b(n-1)]/3,则数列{an+bn}数列{an},{bn}中,a1=3,b1=0,当n≥2时an=[2a(n-1)+b(n-1)]/3bn=[a(n-1)+2b(n-1)]/3则数列{an+bn}的前n项和为?

Question

数列{an},{bn}中,a1=3,b1=0,当n≥2时an=[2a(n-1)+b(n-1)]/3,bn=[a(n-1)+2b(n-1)]/3,则数列{an+bn}
数列{an},{bn}中,a1=3,b1=0,当n≥2时
an=[2a(n-1)+b(n-1)]/3
bn=[a(n-1)+2b(n-1)]/3
则数列{an+bn}的前n项和为?

Answer 1

an=[2a(n-1)+b(n-1)]/3
bn=[a(n-1)+2b(n-1)]/3
相加
an+bn=a(n-1)+b(n-1)
SN=n(a1+b1)=3n