一道数列题求详解,对任意的正整数m,定义f1(m)为m的各位数字的和的平方(例：f1(123)=(1+2+3)2=36),对任意n≥2(n∈N＋),规定fn(m)=f1(fn－1(m)),则f2007(11)=A.16B.49C.169D.256

Question

一道数列题求详解,
对任意的正整数m,定义f1(m)为m的各位数字的和的平方(例：f1(123)=(1+2+3)2=36),对任意n≥2(n∈N＋),规定fn(m)=f1(fn－1(m)),则f2007(11)=
A.16 B.49 C.169 D.256

Answer 1

f2007(11)=f1(f2006(11))=f1(f1(f2005(11)))=f1(f1(f1(f2004(11))))
=...=f1(f1(f1 ...(f1(11)...),一共2007层
=f1(f1(f1 ...(f1(4)...)=f1(f1(f1 ...(f1(16)...)
=f1(f1(f1 ...(f1(49)...)=f1(f1(f1 ...(f1(169)...),从内向外第5层
=f1(f1(f1 ...(f1(256)...),从内向外6层
=f1(f1(f1 ...(f1(169)...),从内向外第7层
=f1(f1(f1 ...(f1(256)...),从内向外第8层
=f1(f1(f1 ...(f1(169)...),从内向外第9层
=...
=f1(f1(256)）,从内向外2006层
=f1(169),从内向外2007层
=（1+6+9）^2=256