>>740

漸化式から
　b[n+1] -1 = (b[n] -1)b[n]
　　= (b[n-1] -1)b[n-1]b[n]
　　= ･･････
　　= (b[1]-1)b[1]b[2]････b[n]
　　= (1/2)a[n],
よって
　(a[n]+1)^2 - Σ[k=2,n+1] (a[k-1])^2
　= 4{b[n+1] - Σ[k=2,n] (b[k] -1)^2} -3
　= 4{b[n+1] + Σ[k=1,n-1] b[k+1] - Σ[k=2,n] (b[k]^2 -b[k] +1)} -3
　= 4{b[2] + Σ[k=2,n] (b[k+1] - b[k]^2 +b[k] -1)} -3
　= 4･b[2] -3,