How do i do this mathematical induction question?

Question

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?

Answer 1

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

Answer 2

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 $n$.

Answer 3

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.