performing induction on $b_{n+1}=1+\sum_{r=1}^{n}b_{r}^{2}$

Question

A sequence is defined by $b_{1}=1$ and $b_{n+1}=b_{n}\left(b_{n}+1\right)$. Prove that for each n: $b_{n+1}=1+\sum_{r=1}^{n}b_{r}^{2}$

Answer 1

Let $n \in \mathbb{N}^\ast$ and $k \in \{1,2...,n-1, n\}$:

We have: $b_{k+1} = b_k^2 + b_k$

Thus: $ b_{k+1} - b_k = b_k^2 $

Therefore: $\sum_{k=1}^{n} (b_{k+1} - b_k) = \sum_{k=1}^{n} b_k^2 $

Therefore: $b_{n+1} - b_1 = \sum_{k=1}^{n} b_k^2 $

Finally: $b_{n+1} = 1 + \sum_{k=1}^{n} b_k^2 $