# 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

##### 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

##### Active 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