Proof by induction

Hi guys, new on tsr, just needed a little help on this proof by induction questions. Just started Further Maths as a year 12 student

I suppose you did the base case?? Did you also write down the assumption?

By assumption, you know that $\displaystyle \sum_{r=1}^k \left( \dfrac{1}{2} \right)^r = 1-\dfrac{1}{2^k}$ so you can use this on your RHS.

Yeah i did the basis step, and the assumption. so i got to this now but im so confused on what to do

You want to 'simplify' this expression and show that it is equal to $1-\dfrac{1}{2^{k+1}}$. I would begin by noting that $\dfrac{1}{2^{k+1}} = \dfrac{1}{2} \cdot \dfrac{1}{2^k}$.

Remember, in the inductive step you wish to deduce that indeed we have $\displaystyle \sum_{r=1}^{k+1} \left( \dfrac{1}{2} \right)^r = 1-\dfrac{1}{2^{k+1}}$

like this???

Well, not what I was going for, but this is valid as well. Your first line is kinda pointless as it adds no value though.

Here you have just used the fact that $\dfrac{1}{2^k} = \dfrac{2}{2\cdot 2^k} = \dfrac{2}{2^{k+1}}$

That said, you can combine the two fractions now quite easily since their denominators are the same.

Ok thank you so much, i added the two fractions. I checked the answer right now and it was right so what would you recommend i do for these types of questions? (like tips)

(I wrote my conclusion too btw)