Maths proof question

How do I show the following expression?:
(1+2+...+r)^2 + (r+1)^3 = (1+2+...+r+r+1)^2

Thank you in advance for any help

Hmm... what tools do you have? Any specific method are you required to use?
Please show some thoughts/workings, even if they are "wrong" or "useless" (they are not).

EDIT: I assume you want to use induction. In that case, please do write down the freebies first, i.e.
(i) Let P(n) be the statement "...",
(ii) [Base step] Show that P(1) is true.
(iii) [Induction assumption] Assume P(k) is true for some k, i.e....
(iv) [Our goal] We want to show that... <- for some reason most people don't write this, even though this is immensely useful.
Then we can talk.

That said, there is a much slicker way without going through all the hassle of induction. Hence why I ask if you are required to use any specific method.

(edited 1 year ago)

Reply 2

davros

Original post by pianowiz

How do I show the following expression?:
(1+2+...+r)^2 + (r+1)^3 = (1+2+...+r+r+1)^2

Thank you in advance for any help

are you allowed to use any previously derived results about sums of series? I can see one obvious line of attack without needing induction...

Reply 3

pianowiz

Original post by davros

are you allowed to use any previously derived results about sums of series? I can see one obvious line of attack without needing induction...

What is the previous line of attack you mean?
It's an induction question and I've managed to get to this point, but I'm not sure how to show how to get from this step to the next.

Reply 4

tonyiptony

Original post by pianowiz

What is the previous line of attack you mean?
It's an induction question and I've managed to get to this point, but I'm not sure how to show how to get from this step to the next.

You'll need to know at least one thing if you want to skip the induction, which I assumed you aren't allowed to:
(i) the explicit expression for the sum of integers from 1 to n (or if you happen to know sum of arithmetic progressions, that will work too)
(ii) just helpful to simplify the algebra, but difference of two squares.

Now, what do you mean by "this point"?
For instance, how much of what I said in #2 did you manage to write down? (I don't care about the actual inductive step for now)

Reply 5

Saeed23

Original post by pianowiz

How do I show the following expression?:
(1+2+...+r)^2 + (r+1)^3 = (1+2+...+r+r+1)^2

Thank you in advance for any help

To prove this expression, we will start with the left-hand side (LHS) of the equation:

LHS = (1+2+...+r)^2 + (r+1)^3

= [r(r+1)/2]^2 + (r+1)^3 (Using the formula for the sum of the first r natural numbers)

= [(r^2 + 2r + 1)/4] + (r+1)^3

= [(r+1)^2/4] + (r+1)^3

= [(r+1)^2][(r+1)/4 + (r+1)]

= [(r+1)^2][(r+2)/2] (Simplifying)

Now, let's look at the right-hand side (RHS) of the equation:

RHS = (1+2+...+r+r+1)^2

= [(r(r+1)/2) + (r+1)]^2 (Using the formula for the sum of the first r natural numbers)

= [(r+1)(r+2)/2]^2

= [(r+1)^2][(r+2)/2]^2

= [(r+1)^2][(r+2)/2] (Simplifying)

Thus, we can see that the LHS and RHS are equal, which proves the given expression:

(1+2+...+r)^2 + (r+1)^3 = (1+2+...+r+r+1)^2

Reply 6

ghostwalker

Original post by Saeed23

...

Please edit/remove your full solution and check the maths forum guidelines - sticky thread towards top of forum.

Reply 7

davros

Original post by Saeed23

To prove this expression, we will start with the left-hand side (LHS) of the equation:

Please remove your solution - it is against posting guidelines!

Please read the forum guidelines here: https://www.thestudentroom.co.uk/showthread.php?t=4919248

Reply 8

davros

Original post by pianowiz

What is the previous line of attack you mean?
It's an induction question and I've managed to get to this point, but I'm not sure how to show how to get from this step to the next.

If the question allowed it, you could use the result for the sum of the 1st n integers. But as per @tonyiptony I would proceed as normal with the inductive hypothesis, base case etc :smile: