Question

数列通向公式a1=1 a2=1 a3=3 a4=5 a5=11

Answer 1

a1=1 a2=1 a3=3 a4=5 a5=11

a2=2*a1-1

a3=2*a2+1

a4=2*a3-1

a5=2*a4+1

a(n+1)=2an+(-1)^n

∴a(n+1)+(-1)^(n+1)/3=2[an+1/3*(-1)^n]

∴[a(n+1)+1/3*(-1)^(n+1)]/[an+1/3*(-1)^n]=2

∴{an+1/3*(-1)^n}为等比数列,公比为2

∴an+1/3*(-1)^n=(1-1/3)*2^(n-1)

∴an=1/3*[2^n-(-1)^n]