1. Using induction, show that the sum of the cubes of the...

1. Using induction, show that the sum of the cubes of the first ( n ) positive integers is given by [ sum_{k=1}^{n} k^{3}=1^{3}+2^{3}+3^{3}+cdots+n^{3}=left(frac{n(n+1)}{2} ight)^{2} ] For full marks you will need to specify all the steps required for a proof by induction. [10 marks]

Transcript

00:01 Problem, using induction, we need to show that the sum of the cubes of the first n positive integers is given by n times n plus 1 divided by 2 whole squared.

00:11 So, first of all, let us consider t of n to be the statement that summation k is equal to 1 to n of k cube is equal to n times n plus 1 over 2 whole squared.

00:26 Now, first of all, let us consider what we call the base step.

00:31 For the base step, we are going to show that this statement holds for the first positive integer, which is 1.

00:39 So for n is equal to 1, what do we have on the left -hand side of this statement? we have the sum of the first 1 positive integer.

00:51 So that's just 1 cube.

00:54 Sorry, the sum of the cube of the first one positive integer.

00:57 So we have one cube, that is one.

01:00 And what do we have on the right hand side? so our formula over here is n times n plus 1 over 2 whole squared.

01:07 So 1 times 1 plus 1 over 2 whole squared.

01:11 So that's 1 times 2 over 2 whole squared.

01:15 That's 1 squared, which is equal to 1.

01:17 And so we can see that the lhs is equal to the rhs.

01:21 And so for n is equal to 1, the statement holds.

01:25 So what can we say? we can say that t of 1 holds.

01:30 T of 1 is true.

01:31 The statement is true for n is equal to 1.

01:34 Our base step is complete.

01:36 Next, consider the induction step.

01:41 Now for the induction step, first of all, what we do is we assume that t of k holds for some positive integer k.

01:56 We're going to assume the t of k holds for that.

01:58 So that just means that 1 cube plus 2 cube up to k cubed.

02:04 This will be equal to k times k plus 1 over 2 whole squared.

02:09 So we assume this.

02:11 And using this, what we want to show is that t of k plus 1 holds.

02:15 So t of k plus 1.

02:17 On the left hand side, we should have 1 cube plus 2 cubed.

02:21 And we go up to k plus 1 cubed.

02:24 And we're going to show that we can obtain this by substituting n is equal to k plus 1 in this formula...

Question

1. Using induction, show that the sum of the cubes of the first ( n ) positive integers is given by [ sum_{k=1}^{n} k^{3}=1^{3}+2^{3}+3^{3}+cdots+n^{3}=left(frac{n(n+1)}{2} ight)^{2} ] For full marks you will need to specify all the steps required for a proof by induction. [10 marks]

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