差成等比数列如何求通项拜托各位大神数列中后一项减前一项成等比,通项怎么求

Question

差成等比数列如何求通项拜托各位大神
数列中后一项减前一项成等比,通项怎么求

Answer 1

就以(an)-(an-1)=K[(an-1)-(an-2)] a1=m；a2=n为例 (an)-(an-1)=K[(an-1)-(an-2)] =K^2[(an-2)-(an-3)] ······=k^(n-2)(a2-a1) (an-1)-(an-2)=K[(an-2)-(an-3)]=··········=k^(n-3)(a2-a1) (an-2)-(an-3)=k[(an-3)-(an-4)]=`````=k^(n-4)(a2-a1) ········ a3-a2=k(a2-a1) a2-a1=a2-a1 将其累加可得an-a1=[k^(n-3)+k^(n-2)······+1](a2-a1) an=[k^(n-2)+k^(n-3)······+1](a2-a1)+a1 an=[k^(n-1)-1]/(k-1)(a2-a1)+a1 an=[k^(n-1)-1]/(k-1)(m-n)+n 最后检验当n=1 和n=2时满足不就行了