Thread: Mathematical induction

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

For

First step

Make

Show

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.

Originally Posted by pickslides

Make

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To be precise, one must define a property, not a function. If , then f(1) is neither true nor false, so one cannot prove f(1). Instead, define P(n) to be and prove P(1) and .

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?

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,....

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



Now you wish to show



Now use the induction hypothesis to get

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

1+5+5^2+...+5^(k+1)=(1+5+5^2+...+5^k)+5^(k+1) (not sure about this line) It's fine 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)



P(k+1)



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



and so