# Math Help - Mathematical induction

1. ## Mathematical induction

I need to use mathematical induction to prove that

1+5+5^2+...+5^n=1/4(5^(n+1)-1)

for all positive intergers n

Thanks

2. For $\displaystyle 1+5+5^2+\dots +5^n=\frac{1}{4}(5^{n+1}-1)$

First step

Make $\displaystyle f(n) = 1+5+5^2+\dots +5^n$

Show $\displaystyle f(1)$

3. Thanks i did the first step, its the second part I am stuck on, the K+1 bit.

If you could offer any help I would be very appreciative.

Thanks.

4. Originally Posted by pickslides
Make $\displaystyle f(n) = 1+5+5^2+\dots +5^n$

Show $\displaystyle f(1)$
To be precise, one must define a property, not a function. If $f:\mathbb{N}\to\mathbb{N}$, then f(1) is neither true nor false, so one cannot prove f(1). Instead, define P(n) to be $\displaystyle 1+5+5^2+\dots +5^n=\frac{1}{4}(5^{n+1}-1)$ and prove P(1) and $\forall k\,(P(k)\to P(k+1))$.

Originally Posted by Arron
its the second part I am stuck on, the K+1 bit.
Why don't you post the induction hypothesis, what you need to prove and what exactly your problem is?

5. This is how far I have got,

P(1) is true, since

1+5^1=6 and 1/4(5^1+1 -1) =6

Now let k>1, and assume that P(k) is true, that is

1+5+5^2+...+5^K = 1/4(5^k+1 -).

We wish to duduce that P(k+1) is true, that is

1+5+5^2+...+5^k+1=1/4(5^K+2 -1)

Now

= (cant work out what goes here)

= 1/4(5^k+2 -1)

Hence P(K) implies P(k+1), for k=1,2....

Hence by mathematical induction, p (n) is true, for n=1,2,....

6. Originally Posted by Arron
This is how far I have got,

P(1) is true, since

1+5^1=6 and 1/4(5^1+1 -1) =6

Now let k>1, and assume that P(k) is true, that is

1+5+5^2+...+5^K = 1/4(5^k+1 -).

We wish to duduce that P(k+1) is true, that is

1+5+5^2+...+5^k+1=1/4(5^K+2 -1)

Now

= (cant work out what goes here)

= 1/4(5^k+2 -1)

Hence P(K) implies P(k+1), for k=1,2....

Hence by mathematical induction, p (n) is true, for n=1,2,....

$\displaystyle \sum_{n=0}^{k}5^n=\frac{1}{4}\left( 5^{k+1}-1\right)$

Now you wish to show $k+1$

$\displaystyle \sum_{n=0}^{k+1}5^n= 5^{k+1}+\sum_{n=0}^{k}5^n$

Now use the induction hypothesis to get

$\displaystyle 5^{k+1}+\frac{1}{4}\left( 5^{k+1}-1\right)=\frac{1}{4}\left( 4\cdot 5^{k+1}+5^{k+1}-1\right)=\frac{1}{2}\left( 5\cdot 5^{k+1}-1\right)=\frac{1}{4}\left( 5^{k+2}-1\right)$

7. Originally Posted by Arron
This is how far I have got,

P(1) is true, since

1+5^1=6 and 1/4(5^1+1 -1) =6

Now let k>1, and assume that P(k) is true, that is

1+5+5^2+...+5^K = 1/4(5^(k+1) -1)

We wish to duduce that P(k+1) is true, that is

1+5+5^2+...+5^(k+1)=1/4(5^(k+2) -1)

Now

= (cant work out what goes here)

= 1/4(5^(k+2) -1)

Hence P(K) implies P(k+1), for k=1,2....

Hence by mathematical induction, p (n) is true, for n=1,2,....

P(k)

$5^0+5^1+5^2+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)$

P(k+1)

$5^0+5^1+5^2+.....+5^{k+1}=\frac{1}{4}\left(5^{k+2}-1\right)$

You should be trying to show that "if" P(k) is true, "then" P(k+1) will also be true.

Proof

If P(k) is true, then

$5^0+5^1+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)$

and so

$5^0+5^1+....+5^k+5^{k+1}=\frac{1}{4}\left(5^{k+1}-1\right)+5^{k+1}$

$=\frac{1}{4}\left(5^{k+1}+(4)5^{k+1}-1\right)=\frac{1}{4}\left((5)5^{k+1}-1\right)$

$=\frac{1}{4}\left(5^{k+2}-1\right)$