You need to show that 1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)((k + 1) + 1) / 2)²
now
Hence P(k+1) is proved true.
Do you need an explanation of part f?
Its part e I am stuck on
Q: Let P(n) be the statement that 1^3 + 2^3 +….n^3 = (n(n + 1) / 2)² for the positive
integer n.
a.) What is the statement P(1)?
b.) Show that P(1), completing the basis step of the proof?
c.) What is the Inductive Hypothesis?
d.) What do you need to prove in the Inductive Step?
e.) Complete the inductive Step?
f.) Explain why these steps show that the formula is true whenever n is a positive integer?
__________________________________________________ ______________________________________________
A:
a.) P(n) = (n(n + 1) / 2)²
P(1) = (1(1 + 1) / 2)²
= (2 / 2)²
= 1
b.) Basis Step:
we will let n = 1
n^3 = (n(n + 1) / 2)²
1^3 = (1(1 + 1) / 2)²
1 = (2 / 2)²
1 = 1
Since both are equal, the basis step hold.
c.) Inductive Hypothesis:
This is the statement 1^3 + 2^3 +….k^3 = (k(k + 1) / 2)²
d.) I have to prove that k > 1 and P(k) implies P(k + 1).
Or.
1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)((k + 1) + 1) / 2)²
1^3 + 2^3 +….k^3 + (k + 1)^3 = (((k + 1)((k + 2)) / 2)²
e.)
1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)((k + 1) + 1) / 2)² + (k + 1)
1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)² · ((k + 2)²) / 4 + (k + 1)
1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)² · (k + 2))² + 4(k + 1)
As per process of induction, we have assumed that P(k) is true.
so we have
1^3 + 2^3 +….k^3 + = (k (k + 1) / 2)² ------ eq1
Now we have to prove that P(k+1 ) is true.
that means we have to prove
1^3 + 2^3 +….k^3 + (k + 1)^3 = ((k + 1)((k + 1) + 1) / 2)²
we have taken the LHS side of above.
then we have used eq1 and simplified the LHS to show that its equal to RHS.
hence we have proved that P(k+1) is true.
Is that clear now?
Now try to explain part f of question. And let us know if you have problem with that.
When doing problems that tends to these 3 steps: Basis Step, Inductive Hypothesis and proving the Inductive Step, I am having difficuly understanding the Inductive Hypotheses. I understand you switch n with (k + 1) but for the prvious problem we added the (k + 1)^3 to the LHS and RHS? How do you know to do that. just like in this problem below:
Q: Prove that 1 · 1! + 2 · 2! + ··· n · n! = (n + 1)! – 1 whenever n is a positive integer?
I have this:
Basis step:
Let n = 1
n · n! = (n + 1)! – 1
1 · 1! = (1 + 1)! – 1
1 · 1 = (2)! – 1
1 = 1
Since they both equal, the basis step holds.
Inductive Hypothesis:
1 · 1! + 2 · 2! + ··· (k + 1) · (k + 1)! = ((k + 1) + 1)! – 1
Would I add a (k + 2)! to both the LHS and RHS?
I think you are very clear with base case. Only conusion is inductive hypothesis and proof.
So here is a bit exlaination.
First we have to assume that statement is true for any number say k, and then we have to prove that statement is correct for next consecutive number k+1.
So if you assume that statement is true for k+1, prove that its correct for k+2.
OR
If you assume that statement is true for k-1, prove tat its true for k
Or similar.
In the first (post 1 of this thread)
we have to prove that 1^3 + 2^3 +….n^3 = (n(n + 1) / 2)²
we have assumed it correct for k.
so we have now
1^3 + 2^3 +….k^3 = (k(k + 1) / 2)² -----------eq1
i.e sum of cube of numbers from 1 to k is (k(k+1)/2)^2
Now we have to prove that statement is true for k+1.
so what do we have to prove?
sum of cubes of number from 1 to k+1 is ((k+1)(k+1+1)/2)^2
Got it?
Now use this sameapproch for any question.
Like in above problem 1! + 2 · 2! + ··· n · n! = (n + 1)! – 1
assume it true for k
so we have assumed 1! + 2 · 2! + ··· k · k! = (k + 1)! – 1
now prove for k+1