Question on induction and the application of an 'equivalent' induction hypothesis.

Question

I am working on the following problem which I decided to solve by induction

Problem: Let $(a_n), (b_n)$ be sequences for $n \geq 1$. Define $B_n:= \sum_{i=1}^n b_n$ for $n \in \mathbb{N}$. Show that: $$\sum_{i=1}^n a_i b_i = a_{n+1}B_n + \sum_{i=1}^n (a_i-a_{i+1})B_i \tag{*} $$

Now to my question. It is easy to show via induction that the base cases are true. I did it for $n=1,2$ and it holds. So I claim (*) to be true for a value $n \in \mathbb{N}$ and take that as my induction hypothesis.

So showing that $n \leadsto n+1 $, I did something that seems to work but I doubt that it's a legitimate operation. $$\sum_{i=1}^{n+1}a_ib_i = \sum_{i=1}^n a_i b_i + a_{n+1}b_{n+1} \overset{*}=a_{n+1}b_{n+1} + a_{n+1} B_n + \sum_{i=1}^n (a_i-a_{i+1})B_i \tag{**} $$

Now I wouldn't know how to proceed, except going for a long calculation of the right sum and hoping that things would simplify at the end, which I think is doable, I just didn't manage to far.

What I did instead was applying $(*)$ again by solving for the sum and substituting it into the expression above (**) above. Meaning that $$(*) \implies \sum_{i=1}^n (a_i - a_{i+1})B_i = \sum_{i=1}^n a_i b_i - a_{n+1} B_n $$

If I substitute this expression into (**) the results follows immediately. However I a daunting feeling that this isn't the right things to do, or is it? How often and in what form am I allowed to make use of the induction hypothesis.

Answer 1

You have to note in (**) that :

$$a_{n+1}b_{n+1} + a_{n+1} B_n = a_{n+1} B_{n+1}$$

Thus (**) is :

$$\sum_{i=1}^{n+1}a_ib_i = \sum_{i=1}^n a_i b_i + a_{n+1}b_{n+1} = a_{n+1} B_{n+1} + \sum_{i=1}^n (a_i-a_{i+1})B_i \tag{***} $$

But we have :

$$\sum_{i=1}^{n+1} (a_i-a_{i+1})B_i = \sum_{i=1}^n (a_i-a_{i+1})B_i + (a_{n+1}-a_{n+2})B_{n+1}$$

from which :

$$\sum_{i=1}^n (a_i-a_{i+1})B_i = \sum_{i=1}^{n+1} (a_i-a_{i+1})B_i - (a_{n+1}-a_{n+2})B_{n+1}$$

Now we substitute into (***) to have :

$$\sum_{i=1}^{n+1}a_ib_i = a_{n+1} B_{n+1} + \sum_{i=1}^{n+1} (a_i-a_{i+1})B_i - (a_{n+1}-a_{n+2})B_{n+1}$$

from which we conclude :

$$\sum_{i=1}^{n+1}a_ib_i = a_{n+2}B_{n+1} + \sum_{i=1}^{n+1} (a_i-a_{i+1})B_i$$