If Φ (r)= 2^r, show that Φ (r+1) =2 Φ(r).
Φ(r)=2r,Φ (r)= 2^r,Φ(r)=2r,
then Φ(r+1)=2r+1=2r⋅21=2⋅2r=2Φ(r),\Phi(r+1)=2^{r+1}=2^r\cdot2^1=2\cdot2^r=2\Phi(r),Φ(r+1)=2r+1=2r⋅21=2⋅2r=2Φ(r),
so the statement Φ(r+1)=2Φ(r)Φ (r+1) =2 Φ(r)Φ(r+1)=2Φ(r) holds.
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