induction q (1 Viewer)

mr.habibbi · Sep 18, 2021

hey all, got a q, doesn't seem too hard but i cant get LHS=RHS in last step

$RTP: 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 =\frac{n(2n^2+1)}{3}$

Life'sHard · Sep 18, 2021

I got it. I couldn't be bothered factorising the LHS. So I ended up just expanding both LHS and RHS of the equation and they equalled.

You should end up with LHS = (k+1)^2 + k^2 + {k(2k^2+1)}/3
I expanded this and it equalled the RHS expanded.

notme123 · Sep 18, 2021

mr.habibbi said:
hey all, got a q, doesn't seem too hard but i cant get LHS=RHS in last step

$RTP: 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 =\frac{n(2n^2+1)}{3}$

Put under common denominator and use long division.

Life'sHard · Sep 18, 2021

notme123 said:
Put under common denominator and use long division.

Long division? It's just a normal induction question. subbing in the 2nd step (n=k) into 3rd step (n=k+1) shouldn't be that difficult. Just don't make any silly mistakes.

mr.habibbi · Sep 18, 2021

try it

Life'sHard · Sep 18, 2021

mr.habibbi said:
try it

I have.

CM_Tutor · Sep 18, 2021

mr.habibbi said:
hey all, got a q, doesn't seem too hard but i cant get LHS=RHS in last step

$RTP: 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 =\frac{n(2n^2+1)}{3}$

Add $n^2$ .

$\begin{align*} \text{IF} \qquad 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 &= \cfrac{n\left(2n^2+1\right)}{3} \\ \text{THEN} \qquad 1^2 + 2^2 + ... + (n-1)^2 + n^2 \color{red} + n^2 \color{black} + (n-1)^2 + ... + 2^2 + 1^2 &=\cfrac{n\left(2n^2+1\right)}{3} + \color{red} + n^2 \color{black} \\ 2\bigg[1^2 + 2^2 + ... + (n-1)^2 + n^2\bigg] &= \cfrac{n}{3}\left(2n^2+1 + 3n\right) \\ &= \cfrac{n}{3}(n+1)(2n + 1) \\ 1^2 + 2^2 + ... + (n-1)^2 + n^2 &= \cfrac{n}{6}(n+1)(2n + 1) \end{align*}$

Now, each step above is reversible and so it is sufficient to prove the standard result, $1^2 + 2^2 + ... + n^2 = \frac{1}{6}n(n+1)(2n + 1)$ , to also prove the required result.

Life'sHard · Sep 18, 2021

CM_Tutor said:
Add $n^2$ .

$\begin{align*} \text{IF} \qquad 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2 &= \cfrac{n\left(2n^2+1\right)}{3} \\ \text{THEN} \qquad 1^2 + 2^2 + ... + (n-1)^2 + n^2 \color{red} + n^2 \color{black} + (n-1)^2 + ... + 2^2 + 1^2 &=\cfrac{n\left(2n^2+1\right)}{3} + \color{red} + n^2 \color{black} \\ 2\bigg[1^2 + 2^2 + ... + (n-1)^2 + n^2\bigg] &= \cfrac{n}{3}\left(2n^2+1 + 3n\right) \\ &= \cfrac{n}{3}(n+1)(2n + 1) \\ 1^2 + 2^2 + ... + (n-1)^2 + n^2 &= \cfrac{n}{6}(n+1)(2n + 1) \end{align*}$

Now, each step above is reversible and so it is sufficient to prove the standard result, $1^2 + 2^2 + ... + n^2 = \frac{1}{6}n(n+1)(2n + 1)$ , to also prove the required result.

I've never seen an induction q be solved like that. I would have just done n=k n=k+1 but I guess this works as well.

mr.habibbi · Sep 18, 2021

Life'sHard said:
I've never seen an induction q be solved like that. I would have just done n=k n=k+1 but I guess this works as well.

do this since ur so smart

CM_Tutor · Sep 18, 2021

But, it can also be done by standard induction. Let's assume that part A, the $n=1$ case, is proven. We then have:

$\textbf{B} \qquad \text{Let $k$ be a value of $n$ for which the result is true. That is:}$

$1^2 + 2^3 + 3^2 + ... + (k-1)^2 + k^2 + (k-1)^2 + ... + 3^2 + 2^2 + 1^2 = \cfrac{k\left(2k^2+1\right)}{3}$

$\text{We must now prove that the result must be true for $n=k+1$. That is, we must prove that:}$

$1^2 + 2^3 + 3^2 + ... + k^2 + (k+1)^2 + k^2 + ... + 3^2 + 2^2 + 1^2 = \cfrac{(k+1)\left[2(k+1)^2+1\right]}{3} \quad \quad \text{. . . . . . . . . . (**)}$

$\begin{align*} \text{LHS}\ &= 1^2 + 2^3 + 3^2 + ... + k^2 + (k+1)^2 + k^2 + ... + 3^2 + 2^2 + 1^2 \\ &= \bigg[1^2 + 2^3 + 3^2 + ... + (k-1)^2 + k^2 + (k-1)^2 + ... + 3^2 + 2^2 + 1^2\bigg] + k^2 + (k+1)^2 \\ &= \cfrac{k\left(2k^2+1\right)}{3} + k^2 + (k+1)^2 \quad \quad \text{using the induction hypothesis (**)} \\ &= \cfrac{1}{3}\left[k\left(2k^2+1\right) + 3k^2 +3(k+1)^2\right] \\ &= \cfrac{1}{3}\left[k\left(2k^2 + 1 + 3k\right) + 3(k+1)^2\right] \\ &= \cfrac{1}{3}\left[k(2k + 1)(k + 1) + 3(k+1)^2\right] \\ &= \cfrac{k+1}{3}\left[k(2k + 1) + 3(k+1)\right] \\ &= \cfrac{k+1}{3}\left(2k^2 + k + 3k + 3\right) \\ &= \cfrac{k+1}{3}\left[2\left(k^2 + 2k + 1\right) + 1\right] \\ &= \cfrac{(k+1)\left[2(k + 1)^2 + 1\right]}{3} \\&= \text{RHS} \end{align*}$

CM_Tutor · Sep 18, 2021

Life'sHard said:
I've never seen an induction q be solved like that. I would have just done n=k n=k+1 but I guess this works as well.

I have the advantage of having seen the standard questions a lot of ties, so I recognise that the question took a standard result and tweaked it. The idea is that shortcuts can save you time to use on other questions, if you spot them, but no one is expected to see all the shortcuts that might exist.

CM_Tutor · Sep 18, 2021

mr.habibbi said:
do this since ur so smart

People posting here, whether it be fellow students, former students, Moderators, etc, do so to get help / learn and to help others. I don't know whether your comment above is directed at me or at @Life'sHard or at someone else, but in any case, I don't appreciate it.

mr.habibbi · Sep 18, 2021

CM_Tutor said:
People posting here, whether it be fellow students, former students, Moderators, etc, do so to get help / learn and to help others. I don't know whether your comment above is directed at me or at @Life'sHard or at someone else, but in any case, I don't appreciate it.

mb just banter with my mate @Life'sHard

CM_Tutor · Sep 18, 2021

mr.habibbi said:
mb just banter with my mate @Life'sHard

Glad to hear it. You might want to bear in mind that tone of voice and the like are not well-conveyed in text form. Your comment evidently read to me as harsh in a way that you didn't intend.

Wizjaro · Sep 18, 2021

mr.habibbi said:
View attachment 32200
do this since ur so smart

How do you prove this?

mr.habibbi · Sep 18, 2021

Wizjaro said:
How do you prove this?

should wait for @Life'sHard solution

Life'sHard · Sep 18, 2021

Wizjaro said:
How do you prove this?

I'm doing tutor hw atm lmao. I'll try and give it a shot afterwards. I've seen this type of question before, so hopefully I can still crack it.

vishnay · Sep 18, 2021

mr.habibbi said:
should wait for @Life'sHard solution

Is this from the hurlstone paper which just requires a butt-load of algebra?

Life'sHard · Sep 18, 2021

vishnay said:
Is this from the hurlstone paper which just requires a butt-load of algebra?

Idk if it's hurlstone or not. But yeah the algebra is kinda nuts if you solve it via mathematical induction wth. Is there another way to solve this? Cause I've had to do induction twice and I'm not seeing the end of the tunnel yet.

vishnay · Sep 18, 2021

Life'sHard said:
Idk if it's hurlstone or not. But yeah the algebra is kinda nuts if you solve it via mathematical induction wth. Is there another way to solve this? Cause I've had to do induction twice and I'm not seeing the end of the tunnel yet.

Check hurlstone mx2 2020 q16 (b) solution

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