Prove via induction $5^0 + 5^1 + 5^2 + \dots 5^n = \frac{5^{n+1}-1}{4}$

Question

Statement:

$$5^0 + 5^1 + 5^2 + \dots 5^n = \frac{5^{n+1}-1}{4}$$

I am having trouble prooving P(k+1) is true. Here is what I have so far:

$$\frac{5^{k+1} -1}{4} + 5^{k+1} = \frac{5^{k+2} -1}{4} $$

LHS $$ \textrm{ stuck here} = \frac{5^{k+1}}{4} - \frac{1}{4} + 5^{k+1} \\\\ OR \\ \textrm{ stuck after this } = \frac{5^{k+1} -1 + 4\cdot 5^{k+1}}{4}$$

No matter how I slice this equation, I am not able to get both sides equal. I have to ask is this even set up properly to begin with? What am I overlooking?

Answer 1

Hint: factor the numerator! I.e. what is $5^{k+1} + 4\cdot 5^{k+1}$?

Answer 2

First, it should be $n$, not $k$. Next factor out $5^{k+1}$ in $$5^{k+1}+4\cdot 5^{k+1}=\cdots\,?$$

Answer 3

Multiplying by $4$ we get $$5^{k+1}-1+4\cdot 5^{k+1}=5^{k+1}-1$$ so it must be $$5^{k+1}+4\cdot 5^{k+1}=5^{k+1}$$ so $$5^{k+1}(1+4)=5^{k+2}$$

Answer 4

$5^{k+1}+4\cdot 5^{k+1}=5\cdot 5^{k+1}=5^{k+2}$.

Answer 5

You are so close !

$$\frac{5^{k+1}-1+4\cdot5^{k+1}}{4} = \frac{5\cdot5^{k+1} -1}{4}.$$