# Thread: Help to prove 5^n+5 < 5^(n+1)

1. ## Help to prove 5^n+5 < 5^(n+1)

Prove 5^n+5 < 5^(n+1) for al n elements of N

So i started this by using induction and used n=1 for my base case which i got 10<25 which is true.

Then i assumed that 5^k+5<5(k+1) for all k elements of N for my induction hypothesis and computed the following as my induction step:

5(k+1)+5< 5^(k+1)+1
soo i tried to split the right side to (5^k)x(5^2) then i got stuck
Can anyone please help me ? Im not sure how to make the right side equal 5^(k+1)

2. Originally Posted by jess0517
Prove 5^n+5 < 5^(n+1) for al n elements of N

So i started this by using induction and used n=1 for my base case which i got 10<25 which is true.

Then i assumed that 5^k+5<5(k+1) for all k elements of N for my induction hypothesis and computed the following as my induction step:

5(k+1)+5< 5^(k+1)+1
soo i tried to split the right side to (5^k)x(5^2) then i got stuck
Can anyone please help me ? Im not sure how to make the right side equal 5^(k+1)

$5^{k+1}+5=5\cdot 5^k+5=4\cdot 5^k+\left(5^k+5)<4\cdot 5^k+5^{k+1}<5\cdot 5^k+5^{k+1}=2\cdot 5^{k+1}<5\cdot 5^{k+1}=5^{k+2}$

Tonio

3. Wow. Amazing how many of those inequalities are not tight, and yet you still get the result. The original inequality must be very loose indeed!

4. Originally Posted by tonio
$5^{k+1}+5=5\cdot 5^k+5=4\cdot 5^k+\left(5^k+5)<4\cdot 5^k+5^{k+1}<5\cdot 5^k+5^{k+1}=2\cdot 5^{k+1}<5\cdot 5^{k+1}=5^{k+2}$

Tonio
Thank you soo much this was great help.. im just a tad confused how the 4x5^k and the 2x5^k+1 got there.

5. Thread closed due to this member deleting questions after getting help.