How to show $\sum^{n}_{k=0}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by induction?

Question

Let $n\in\mathbb{N}$ and $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_{n+1}\in\mathbb{R}$ and $A_k=\sum^{n}_{k+1}{a_j}$ with $1\leq k\leq n$.

Show that $\sum^{n}_{k=1}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by induction.

I tried several times, however, I have a hard time to resolve this in the induction step. This is what I tried so far and looks most promising to me:

$$\sum^{n+1}_{k=1}{a_kb_k}=a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k}$$ $$=a_{n+1}b_{n+1}+\left(A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}\right)$$ $$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)(b_k-b_{k+1})}\right)$$ $$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)(b_k-b_{k+1})}\right)$$ $$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_k}-\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_{k+1}}\right)$$ $$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+b_1a_1+\sum^{n}_{k=2}{\left(\sum^{k}_{i=1}{a_i}\right)b_k}-b_{n+1}\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)}-\sum^{n-1}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)b_{k+1}}\right)$$ $$=a_{n+1}b_{n+1}+\left(\left(\sum^{n}_{k=1}{a_k}\right)b_{n+1}+b_1a_1-b_{n+1}\sum^{n}_{k=1}{\left(\sum^{k}_{i=1}{a_i}\right)}\right)$$ $$=a_{n+1}b_{n+1}+\left(b_1a_1+b_{n+1}\left(A_n-\sum^{n}_{k=1}A_k\right)\right)$$ $$=???$$

I'd appreciate any hints or spotted errors!

Answer 1

Posting my answer from chat:

$$ \begin{align} \sum^{n+1}_{k=1}{a_kb_k} &= a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k} \\ &= a_{n+1}b_{n+1}+A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})} \\ &= a_{n+1}b_{n+1}+A_nb_{n+1} - A_{n+1} (b_{n+1}-b_{n+2})+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\ &= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - (A_{n+1} - A_n)b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\ &= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - a_{n+1}b_{n+1}+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\ &= A_{n+1}b_{n+2} + \sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \end{align} $$

Answer 2

It is not a good idea here to split up the term $A_n$. I would suggest instead: $$ a_{n+1}b_{n+1} + \left[A_n b_{n+1} + \sum_{k=1}^n A_k(b_k-b_{k+1}) \right] = A_{n+1} b_{n+1} + \sum_{k=1}^n A_k(b_k-b_{k+1}) = A_{n+1} b_{n+2} + \sum_{k=1}^{n+1} A_k(b_k-b_{k+1})$$

Answer 3

Hint:

If it must be shown that $$\sum_{k=0}^nr_k=p_n+\sum_{k=1}^nq_k$$ then this can be done by showing that $r_0+r_1=p_1+q_1$ and secondly $r_{n+1}=p_{n+1}-p_n+q_{n+1}$