# induction question (1 Viewer)

#### tutor01

##### Member
Can anyone prove by induction that

n(x+a)^(n-1) = sum_{r=0}^{n} r . nCr . x^(r-1) . a^(n-r) for n=1,2,3,...

Thanks.

#### ultra908

##### Active Member
Uh can you send a photo or write in Latex because your equation is a bit hard to understand.

#### Drdusk

##### π
Moderator
Uh can you send a photo or write in Latex because your equation is a bit hard to understand.
$\bg_white \text{Prove}$
$\bg_white n(x+a)^{n-1} = \sum_{r = 0}^{n}r{n\choose r}x^{r-1}a^{n-r}$

#### tutor01

##### Member
$\bg_white \text{Prove}$
$\bg_white n(x+a)^{n-1} = \sum_{r = 0}^{n}r{n\choose r}x^{r-1}a^{n-r}$
Yes that's the one.

#### harrowed2

##### Member
Here is my working for that question. I have left the n=1 step out to allow the solution to fit on one page.

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#### Drongoski

##### Well-Known Member
Love your very nice handwriting too, harrowed2.

Thank you!

#### CM_Tutor

##### Well-Known Member
Unless required to prove this by induction, this is an excellent example of where an alternative method is much easier.

This result is the direct application of differentiation to the binomial theorem.

The binomial theorem states that $\bg_white (x+a)^n = \sum_{r=0}^n {^n C_r} a^{n - r} x^r = {^n C_0} a^n + {^n C_1} a^{n - 1} x + {^n C_2} a^{n - 2} x^2 + ... {^n C_n} x^n$

Differentiating with respect to x yields: $\bg_white n (x+a)^{n-1} \times 1 = 0 + {^n C_1} a^{n - 1} + {^n C_2} a^{n - 2} 2x + ... {^n C_n} nx^{n-1} = \sum_{r=1}^n {^n C_r} a^{n - r} r x^{r-1}$

Noting the term in the sum is zero when $\bg_white r = 0$, we have: $\bg_white n (x+a)^{n-1} = \sum_{r=0}^n r {^n C_r} x^{r-1} a^{n - r}$ as required

Something to bear in mind here: Questions sometimes use the word "otherwise", as in "Using induction or otherwise, prove ...". The word "otherwise" pretty much always indicates that either there is an alternative method that is much quicker, or that there is an alternative approach that will lead to a huge mess and you should avoid it. A question with a preferable "otherwise" approach offers the chance to pick up time in an exam, which you can then use on other questions. It is worth stopping to consider when the word "otherwise" appears and asking if a better way is available.

#### harrowed2

##### Member
Unless required to prove this by induction, this is an excellent example of where an alternative method is much easier.

This result is the direct application of differentiation to the binomial theorem.

The binomial theorem states that $\bg_white (x+a)^n = \sum_{r=0}^n {^n C_r} a^{n - r} x^r = {^n C_0} a^n + {^n C_1} a^{n - 1} x + {^n C_2} a^{n - 2} x^2 + ... {^n C_n} x^n$

Differentiating with respect to x yields: $\bg_white n (x+a)^{n-1} \times 1 = 0 + {^n C_1} a^{n - 1} + {^n C_2} a^{n - 2} 2x + ... {^n C_n} nx^{n-1} = \sum_{r=1}^n {^n C_r} a^{n - r} r x^{r-1}$

Noting the term in the sum is zero when $\bg_white r = 0$, we have: $\bg_white n (x+a)^{n-1} = \sum_{r=0}^n r {^n C_r} x^{r-1} a^{n - r}$ as required

Something to bear in mind here: Questions sometimes use the word "otherwise", as in "Using induction or otherwise, prove ...". The word "otherwise" pretty much always indicates that either there is an alternative method that is much quicker, or that there is an alternative approach that will lead to a huge mess and you should avoid it. A question with a preferable "otherwise" approach offers the chance to pick up time in an exam, which you can then use on other questions. It is worth stopping to consider when the word "otherwise" appears and asking if a better way is available.
Yes, the question came from the new Maths in Focus Ext 2 textbook in ch 5 Mathematical Induction Exercise 5.04 Q3 part b.

#### Pedro123

##### Member
Heads up - using the method of differentiation (Which is significantly easier) and induction for such problems is quite common in a lot of 3 unit exams, so just make sure you are aware of that process

#### 5uMath

##### Member
Yes, the question came from the new Maths in Focus Ext 2 textbook in ch 5 Mathematical Induction Exercise 5.04 Q3 part b.
Would you be able to share a copy of that textbook, perhaps online through google drive? I know I'm late ahaha, but I'm really limited on resources.

#### harrowed2

##### Member
Heads up - using the method of differentiation (Which is significantly easier) and induction for such problems is quite common in a lot of 3 unit exams, so just make sure you are aware of that process
Yes, I would have normally have solved this question with differentiation, but the textbook asked for the statement to be proved using Mathematical Induction, and was a "hence prove" follow on from part a.

#### 5uMath

##### Member
Yes, I would have normally have solved this question with differentiation, but the textbook asked for the statement to be proved using Mathematical Induction, and was a "hence prove" follow on from part a.
Would you say the new maths in focus for 4u is good?

#### Drdusk

##### π
Moderator
Would you say the new maths in focus for 4u is good?
Maths in focus was never good. I haven't seen the book but I'm 99.9% sure it's trash like all the other ones.

The only thing it's 'sort of'? good for is explaining concepts from what I hear through other students.

#### ultra908

##### Active Member
Would you say the new maths in focus for 4u is good?
i'd say the difficulty level isn't up to cambridge or fitzpatrick. Cambridge exercises have 3 levels of difficulty and maths in focus ends at like the first level lol.

#### 5uMath

##### Member
i'd say the difficulty level is up to cambridge or fitzpatrick.
Yeah, I have fitzpatrick - pretty basic tbh. Ive found a cambridge book so ill see how that goes

#### CM_Tutor

##### Well-Known Member
Maths in focus was never good. I haven't seen the book but I'm 99.9% sure it's trash like all the other ones.

The only thing it's 'sort of'? good for is explaining concepts from what I hear through other students.
It's my understanding that the new syllabus Maths in Focus is better than the older ones written by Margaret Grove... but I agree that the old books were not good books.

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#### ultra908

##### Active Member
Yeah, I have fitzpatrick - pretty basic tbh. Ive found a cambridge book so ill see how that goes
i meant isnt lol

#### 5uMath

##### Member
i meant isnt lol
Oh ahaha. The questions are usually like Q14 or 15 of HSC, so not that bad. His chapter reviews are very easy at times, especially for vectors

#### harrowed2

##### Member
Would you say the new maths in focus for 4u is good?
It's ok, does some topics well, but has some material that is not covered by the others, which adds to the confusion of what to concentrate on. My pick is the new Cambridge text, followed by Fitzpatrick Ed 3 and the new Terry Lee.