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My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$

1. So first step i have to prove LHS = RHS when $n=1$, which is true.


2. Then I assume the statement is true for $n=k$
   


3. Since the statement is true for $n=k$ then for $n=k+1$

My workings:

$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$

LHS: $5(2^{k-1}) + 5(2)^k$

Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?

$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$

For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$

So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.

Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.