2

"Prove using simple induction that for each integer $n \geq 1$, $$ 5 + 5^2 + 5^3 +..... + 5^n = \frac{5^{n+1}-5}4 $$

so I start with base step

base step: $n = 1$ $$5^1 = \frac{5^{1+1}-5}4 $$ is correct.

induction step: so we assume that $P(k)$ is true for all integers $≥1$ $$ 5^{(k+1)} = \frac{5^{(k+1)+1}-5}4 $$ this is where I get stuck :X, i'm confused about the "$5 + 5^2 + 5^3 +\ldots$ " notation in question and just generally where to go from here

Henrik supports the community
  • 6,831
discord1
  • 43
  • See http://math.stackexchange.com/questions/658992/proving-the-geometric-sum-formula-by-induction and https://proofwiki.org/wiki/Sum_of_Geometric_Progression#Induction_Step – lab bhattacharjee Apr 02 '16 at 09:18
  • 6
    http://math.stackexchange.com/questions/1426699/double-induction-example-1-q-q2-q3-cdots-qn-1-qn-frac/1426799#1426799 – Mastrem Apr 02 '16 at 09:19
  • You assume it is true for $n=k $ and if it is then show $n=k+1$ must also be true. – Karl Apr 02 '16 at 09:52

1 Answers1

4

You have assumed that the statement is true for all $n$ uptill $k$ and now you want to prove the statement for $n = k+1$

For $n = k+1$ the summation on the LHS is

$$5 + 5^2 + \dots + 5^k + 5^{k+1}$$

We want to prove that this is equal to $\frac{5^{k+2}-5}{4}$.

Let's simplify the LHS using our induction hypothesis that the statement is true for $n = k$. We have

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

Using the above equation below we get

$$\underbrace{5 + 5^2 + \dots + 5^k}_{\text{Use induction hypothesis}} + 5^{k+1} = \frac{5^{k+1}-5}{4} + 5^{k+1}$$ $$ = \frac{5^{k+1}-5 + 4\times5^{k+1}}{4} $$ $$ = \frac{5\times 5^{k+1}-5}{4} = \frac{5^{k+2}-5}{4}$$

This is the same as the RHS of out statement.

Q.E.D.

Banach Tarski
  • 2,207
  • thanks, that makes a lot of sense. but what do you mean where you underlined "Use induction hypothesis", didnt we already do that when we assumed n = k? – discord1 Apr 03 '16 at 08:47