Please could anyone help!
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
To be precise, one must define a property, not a function. If $\displaystyle f:\mathbb{N}\to\mathbb{N}$, then f(1) is neither true nor false, so one cannot prove f(1). Instead, define P(n) to be $\displaystyle \displaystyle 1+5+5^2+\dots +5^n=\frac{1}{4}(5^{n+1}-1)$ and prove P(1) and $\displaystyle \forall k\,(P(k)\to P(k+1))$.
Why don't you post the induction hypothesis, what you need to prove and what exactly your problem is?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
1+5+5^2+...+5^k+1=(1+5+5^2+...+5^k)+(5^k+1) (not sure about this line)
= (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,....
So your induction hypothesis states
$\displaystyle \displaystyle \sum_{n=0}^{k}5^n=\frac{1}{4}\left( 5^{k+1}-1\right)$
Now you wish to show $\displaystyle k+1$
$\displaystyle \displaystyle \sum_{n=0}^{k+1}5^n= 5^{k+1}+\sum_{n=0}^{k}5^n$
Now use the induction hypothesis to get
$\displaystyle \displaystyle 5^{k+1}+\frac{1}{4}\left( 5^{k+1}-1\right)=\frac{1}{4}\left( 4\cdot 5^{k+1}+5^{k+1}-1\right)=\frac{1}{2}\left( 5\cdot 5^{k+1}-1\right)=\frac{1}{4}\left( 5^{k+2}-1\right)$
P(k)
$\displaystyle 5^0+5^1+5^2+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)$
P(k+1)
$\displaystyle 5^0+5^1+5^2+.....+5^{k+1}=\frac{1}{4}\left(5^{k+2}-1\right)$
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
$\displaystyle 5^0+5^1+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)$
and so
$\displaystyle 5^0+5^1+....+5^k+5^{k+1}=\frac{1}{4}\left(5^{k+1}-1\right)+5^{k+1}$
$\displaystyle =\frac{1}{4}\left(5^{k+1}+(4)5^{k+1}-1\right)=\frac{1}{4}\left((5)5^{k+1}-1\right)$
$\displaystyle =\frac{1}{4}\left(5^{k+2}-1\right)$