Prove by mathematical induction for every natural number n. $5+25+125+\cdots+5^n=5/4(5^n-1)$

Question

There's one thing I don't understand. In the work shown for this problem in the image below, why is it adding $5^{k+1}$ to both sides?

https://i.stack.imgur.com/SYqzf.jpg (Part 1)

https://i.stack.imgur.com/t1e3J.jpg (Part 2)

Shouldn't that step be this instead? $5/4(5^k-1)+5^{k+1}=5/4(5^{k+1}-1)$

Answer 1

That's how a proof by recurrence works: assuming that the statement is true for $k$, show that the statement is also true for $k+1$. To complete the proof we then need to check by hand that the statement is true for some starting $k$, usually $k=1$ or $k=2$.

Here, we assume that $$5+25+\cdots+5^k=(5/4)(5^{k}-1).$$ From that assumption, we can write $$5+25+\cdots+5^k+5^{k+1}=(5/4)(5^{k}-1)+5^{k+1}.$$ Then, after some simplifications, we find that $$(5/4)(5^k-1)+5^{k+1}=(5/4)(5^{k+1}-1)$$ which is exactly the result we wanted to get.

Summing it up, we have shown that if $5+\cdots+5^{k}=(5/4)(5^{k}-1)$, then $5+\cdots+5^{k+1}=(5/4)(5^{k+1}-1)$.

Is this more clear?

Answer 2

The base step is clear. Suppose it holds for $n-1$; then, by hypothesis $$ 1+\dots+5^{n-1}=\frac{5}{4}(5^{n-1}-1) $$ Therefore $$ 1+\dots+5^{n-1}+5^n=\frac{5}{4}(5^{n-1}-1)+5^n= \frac{1}{4}(5^n-5+4\cdot 5^n) $$ Can you go on from here?

Answer 3

Although this problem is solvable by induction, here is a simpler solution :

Let $S = 5 + 25 + 125 + ... + 5^n$. Then $5S = 25 + 125 + ... + 5^n + 5^{n + 1}$. So, $4S = 5S - S = 5^{n + 1} - 5$. Thus, $S = \frac{5^{n + 1} - 5}{4} = \frac{5(5^n - 1)}{4}$. QED