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

• July 12th 2010, 04:38 PM
jess0517
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)
• July 12th 2010, 06:12 PM
tonio
Quote:

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
• July 12th 2010, 06:20 PM
Ackbeet
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!
• July 12th 2010, 06:36 PM
jess0517
Quote:

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.
• July 14th 2010, 02:24 AM
mr fantastic
Thread closed due to this member deleting questions after getting help.