This identity should be very hard to prove - because it's not true!
Counterexample: consider .
can anyone help with this induction problem:
n³ = ¼ n² (n+1)²
i'm having trouble proving true for n = k+1
i need to prove that (k+1)³ = ¼ (k+1)² (k+2)² (i hope this is right)
by assumption the LHS should be ¼ k² (k+1)² (k+1) is it not?
can someone help me as to solving this?
Hello, Chuck!
You've already verified , right?Prove by induction: .
We assume is true: .
Add to both sides:
. .
. . The left side is the left side of
Factor the right side: .
. . This is the right side of
Therefore, we have: .
. . and the inductive proof is complete.
I was going to start a new thread but I decided to tagg along this one. Okay I have two problems. Please explain this clearly. Im really having a hard time understanding this stuff.
Prove by induction that for any x, if x 1, then 3n 1 + 2n.
Step 1:For x =, P() = [(3 × 1) = 1 + 2].The base case is (Type T/F.).Step 2:For k suppose that 3k 1 + 2k is(Type T/F.)Step 3:Add in the termNow the expression is(k + 1) 1 + 2(k + 1)This becomes3k + (1 + 2) + 2We conclude the statement is(Type T/F.)because the left side increases more than the right.
Prove by induction that for any x, if x 1, then 5n 1 + 4n .
Step 1:For x = 1, P(1) = [(5 × 1) = 1 +].The base case is (Type T/F.).Step 2:For k suppose that 5k 1 + 4k is(Type T/F.)Step 3:Add in the termNow the expression is(k + 1) 1 +(k + 1)This becomes5k + (1 + 4) + 4We conclude the statement is (Type T/F.)because the left side increases more than the right.
Thank You!
ok, so first of all, you had different variables, everything must be in one variable. you cant make a statement about x and then prove something about n when we don't know anything about n. i changed everything to x. Here's how i would do them
Let P(x): "If , then for all integers "
We have for the base case:
P(1): , so P(1) is true
Assume P(k) is true for some k 1. We show that P(k + 1) is true
Since P(k) is true, we have:
It follows that:
, since we add 3 to the left and 2 to the right
But we can factor this so that:
which is P(k + 1)
Thus the inductive proof is complete and P(x) holds true for all integers x 1
This one is pretty much the same as the last.Prove by induction that for any n, if n 1, then 5n 1 + 4n .
Let P(n): "If , then for all integers "
So P(1): . So P(1) is true
Assume P(k) true for some k 1, we show that P(k + 1) is true
Since P(k) is true, we have:
It follows that:
Factoring we get:
, which is P(k + 1)
Thus the inductive proof is complete, and P(n) holds for all integers n 1