Proof by Induction

An outline of steps to get to the right answer will be:

Step 1: Show that it's true for

n=1

;

Step 2: Note that

\displaystyle \sum_{r=1}^{k+1} r.3^{r-1} = \left( \sum_{r=1}^k r.3^{r-1} \right) + (k+1).3^k

and apply the induction hypothesis to the big bracket;

Step 3: Rearrange stuff.

If you're still stuck, show your working and we'll see where you're going wrong.

[For typing maths into TSR, look at the Guide to LaTeX.]

(edited 11 years ago)

Actually it isn't even true. Where did you get this from?

Reply 5

raheem94

13

Original post by nuodai

An outline of steps to get to the right answer will be:

Step 1: Show that it's true for

n=1

;

Step 2: Note that

\displaystyle \sum_{r=1}^{k+1} r^3(r-1) = \left( \sum_{r=1}^k r^3(r-1) \right) + (k+1)^3((k+1)-1)

and apply the induction hypothesis to the big bracket;

Step 3: Rearrange stuff.

If you're still stuck, show your working and we'll see where you're going wrong.

[For typing maths into TSR, look at the Guide to LaTeX.]

The OP has written the question as: Sum of (from 1 to n) r[3^(r-1)]=¼[3^n(2n-1)+1]:

It means

\displaystyle \sum _{r=1}^n r(3^{r-1})= ...

Original post by raheem94

The OP has written the question as: Sum of (from 1 to n) r[3^(r-1)]=¼[3^n(2n-1)+1]:

It means

\displaystyle \sum _{r=1}^n r(3^{r-1})= ...

He has now but he hadn't then!

Reply 7

OP

Hi guys, thanks for the replies, Raheem is right, the question is as he stated, sorry i could'nt put it in the normal form.
Thanks

And yes, it seems unbelievable, but I did originally type the question wrongly, sorry

Reply 8

Original post by raheem94

The OP has written the question as: Sum of (from 1 to n) r[3^(r-1)]=¼[3^n(2n-1)+1]:

It means

\displaystyle \sum _{r=1}^n r(3^{r-1})= ...

The OP edited their post whilst I was writing my reply. I'll edit my post appropriately - thanks.

Reply 9

OP

I get to =¼[3^k(2k-1)+1+4(k+1)3^k]

Then I cannot continue

Reply 10

There's actually a sneaky way to do this question.

Let

f(x)=x^r

, then

f'(x)=rx^{r-1}

and so

f'(3)=r.3^{r-1}

.

So extending this to the sum we have

\displaystyle \sum_{r=1}^n r.3^{r-1} = \left[ \dfrac{d}{dx} \sum_{r=1}^n x^r \right]_{x=3}

The sum in the brackets is a geometric series which evaluates to

\dfrac{x^{n+1}-1}{x-1}

, which differentiates (by the quotient rule) to give

\dfrac{(n+1)x^n(x-1) - (x^{n+1}-1)}{(x-1)^2}

. Plugging in

x=3

and simplifying gives the desired result.

[But you've been asked to prove this by induction.]

(edited 11 years ago)

Reply 11

Original post by helpstoask

I get to =¼[3^k(2k-1)+1+4(k+1)3^k]

Then I cannot continue

You're close, you just need to simplify. The +1 is fine because that's in the final answer; so take out

3^k

as a common factor from the other bits and see if you can rearrange to get the required form.

Original post by helpstoask

I get to =¼[3^k(2k-1)+1+4(k+1)3^k]

Then I cannot continue

That is correct.

Expand it all out then collect up like terms.

Factorise

3^k

and remember that

3 \times 3^k = 3^{k+1}

Reply 13

OP

Still cannot, finish it, I don't know why, aiming for : ¼ [3^k+1 (2k+1) +1]
Sorry

Reply 14

Original post by helpstoask

Still cannot, finish it, I don't know why, aiming for : ¼ [3^k+1 (2k+1) +1]
Sorry

Right, so you have

\dfrac{1}{4}\big[3^k(2k-1)+ 1+4(k+1)3^k \big]

Take out the

3^k

as a common factor to get

\dfrac{1}{4}\big[ 3^k[2k-1 + 4(k+1)] + 1\big]

Can you see what to do now, given what you've been told so far?

Hint: simplify the bracket after

3^k

and then look for common factors.

(edited 11 years ago)

Original post by helpstoask

Still cannot, finish it, I don't know why, aiming for : ¼ [3^k+1 (2k+1) +1]
Sorry

That's what you get if you follow my advice.

Reply 16

raheem94

13

Original post by helpstoask

I get to =¼[3^k(2k-1)+1+4(k+1)3^k]

Then I cannot continue

\displaystyle \frac14 (3^k(2k-1)+1+4(k+1)3^k ) = \frac14 ((2k) 3^k - 3^k + 1 + (4k)3^k+(4)3^k ) = \frac14 (3^k(2k) + 3.3^k +(4k)3^k +1)= \frac14 (3^k(2k) +(4k)3^k + 3^{k+1} +1) = \frac14 (3^k(6k) + 3^{k+1} +1)

Now try to simplify,

3^k(6k) \ \ to \ \ 2k(3^{k+1})

(edited 11 years ago)

Reply 17

bong

Original post by nuodai

You're close, you just need to simplify. The +1 is fine because that's in the final answer; so take out

3^k

as a common factor from the other bits and see if you can rearrange to get the required form.

but how do you factorise out the 3^k when there is a +1 in that first term?

Original post by nuodai

Right, so you have

\dfrac{1}{4}[3^k(2k-1)+1+4(k+1)3^k]

Take out the

3^k

as a common factor to get

\dfrac{1}{4}[3^k(2k-1 + 4(k+1)) + 1]

Can you see what to do now, given what you've been told so far?

Hint: simplify the bracket after

3^k

and then look for common factors.

I suggested expanding it all for a reason, students with less ability than you struggle with anything but the simplest factorisation.

(edited 11 years ago)

Reply 19