Proof by induction question

Image and question below

The first term should be (-1)^k, not just (-1). Then factorise out
1/2(-1)^(k+1) (k+1)
from both terms and simplify.

Ohh. Why (-k…. though? Yes, I understand that 1/2(-1)^(k+1) (k+1) multiplied by -k, would give the original 1/2(-1)^k+1(k+1), but exactly how would it give it?

Maybe as a passing comment:

I don't like how the marking scheme has written the solution. Aside from the the lack of words, which seems to be something I complain a lot about, the key ingredient in induction is to show where you actually use the induction assumption by literally writing the phrase "by induction assumption" on the correct line (or if you're lazy like me, write "I.A." on the correct equal sign), not at the end.

Also, I tend to start the induction step with "we would like to show that 'blah'", just to know what the goal is.
Sometime (and in fact more often than not) you could get lost in a bunch of calculations that you just forgot what you're trying to achieve. Writing down your goal often gives you a hint as to what your next logical step should be.

For instance in this case, you have in the induction step:
$\displaystyle\sum^{k+1}_{r=1}(-1)^{r}r^2 =\sum^{k}_{r=1}(-1)^{r}r^2 + (-1)^{k+1}(k+1)^2 \\ \stackrel{I.A.}{=} \frac{1}{2}(-1)^{k}k(k+1)+ (-1)^{k+1}(k+1)^2$

If we know explicitly what your goal is, i.e. $\displaystyle...=\frac{1}{2}(-1)^{k+1}(k+1)(k+2)$, there is a strong hint that factorizing is a good idea. Of course now it comes down to how good are you at factorizing expressions... (again, borrowing from calculus, people often say calculus is not hard, it's the algebra)

If we know explicitly what your goal is, i.e. $\displaystyle...=\frac{1}{2}(-1)^{k+1}(k+1)(k+2)$, there is a strong hint that factorizing is a good idea. Of course now it comes down to how good are you at factorizing expressions... (again, borrowing from calculus, people often say calculus is not hard, it's the algebra)

Not sure what you mean. You want a (-1)^(k+1) and (-1)^k = -1*(-1)^(k+1).
Strictly you divide by -1, but thats obv the same.

Maybe as a passing comment:

I don't like how the marking scheme has written the solution. Aside from the the lack of words, which seems to be something I complain a lot about, the key ingredient in induction is to show where you actually use the induction assumption by literally writing the phrase "by induction assumption" on the correct line (or if you're lazy like me, write "I.A." on the correct equal sign), not at the end.

Also, I tend to start the induction step with "we would like to show that 'blah'", just to know what the goal is.
Sometime (and in fact more often than not) you could get lost in a bunch of calculations that you just forgot what you're trying to achieve. Writing down your goal often gives you a hint as to what your next logical step should be.

For instance in this case, you have in the induction step:
$\displaystyle\sum^{k+1}_{r=1}(-1)^{r}r^2 =\sum^{k}_{r=1}(-1)^{r}r^2 + (-1)^{k+1}(k+1)^2 \\ \stackrel{I.A.}{=} \frac{1}{2}(-1)^{k}k(k+1)+ (-1)^{k+1}(k+1)^2$

If we know explicitly what your goal is, i.e. $\displaystyle...=\frac{1}{2}(-1)^{k+1}(k+1)(k+2)$, there is a strong hint that factorizing is a good idea. Of course now it comes down to how good are you at factorizing expressions... (again, borrowing from calculus, people often say calculus is not hard, it's the algebra)

I’ll show my question visually so maybe you can understand what I mean:

I’ll show my question visually so maybe you can understand what I mean:

a^k * a/a * #
a^(k+1) * #/a
You have your a^(k+1) factor and whats left is #/a.

It all made sense, until this part and specifically the bolded part

I thought that, if you multiplied a^k * a/a * #, it would give you a^k+1 * #, because you only need to multiply a term once with another term. For example, when you do 2 x 3 x 4, it becomes 6 x 4, not 6 x 8. Why why the #/a (bolded part)?

12 = 3*4 = 3*(3/3)*4 = (3^2)*(4/3)
Youre effectively multiplying by 1 = 3/3. You use the *3 to increase the power of the first term, then you must divide the second term by 3.

Thank you! One question, the brackets are basically useless right? Because if I had these brackets in mind, I wouldn’t have been able to realise that k becomes -k, given that k was attached to the k+1 via the “[ ]” brackets. Are the brackets just for convention, or do they actually serve a purpose? (which would be confusing if they do, because again, I wouldn’t have been able to recognise that k also becomes affected despite being attached to the k+1 via the “[ ]” brackets)