作业帮斐波那契数列的求和公式
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/05/16 17:24:01
斐波那契数列的求和公式
斐波那契数列的通项公式为
an=√5/5[(1+√5)/2]^n-√5/5[(1-√5)/2]^n,设bn=√5/5[(1+√5)/2]^n,cn=√5/5[(1-√5)/2]^n
则an=bn-cn,{bn}是公比为(1+√5)/2的等比数列,{cn}是公比为(1-√5)/2的等比数列,
bn的前n项和Bn=√5/5[(1+√5)/2]*(1-[(1+√5)/2]^n)/(1-[(1+√5)/2])
=(3√5+5)([(1+√5)/2]^n-1)/10
cn的前n项和Cn=√5/5[(1-√5)/2]*(1-[(1-√5)/2]^n)/(1-[(1-√5)/2])
=(3√5-5)([(1-√5)/2]^n-1)/10
所以an的前n项和An=a1+a2+…+an=b1-c1+b2-c2+…+bn-cn=Bn-Cn
=(3√5+5)([(1+√5)/2]^n-1)/10-(3√5-5)([(1-√5)/2]^n-1)/10
={(3√5+5)([(1+√5)/2]^n-1)-(3√5-5)([(1-√5)/2]^n-1)}/10
an=√5/5[(1+√5)/2]^n-√5/5[(1-√5)/2]^n,设bn=√5/5[(1+√5)/2]^n,cn=√5/5[(1-√5)/2]^n
则an=bn-cn,{bn}是公比为(1+√5)/2的等比数列,{cn}是公比为(1-√5)/2的等比数列,
bn的前n项和Bn=√5/5[(1+√5)/2]*(1-[(1+√5)/2]^n)/(1-[(1+√5)/2])
=(3√5+5)([(1+√5)/2]^n-1)/10
cn的前n项和Cn=√5/5[(1-√5)/2]*(1-[(1-√5)/2]^n)/(1-[(1-√5)/2])
=(3√5-5)([(1-√5)/2]^n-1)/10
所以an的前n项和An=a1+a2+…+an=b1-c1+b2-c2+…+bn-cn=Bn-Cn
=(3√5+5)([(1+√5)/2]^n-1)/10-(3√5-5)([(1-√5)/2]^n-1)/10
={(3√5+5)([(1+√5)/2]^n-1)-(3√5-5)([(1-√5)/2]^n-1)}/10