Series Question

[upbeat prism]

can't answer part B using the equations from part A

[lofty cipher]

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[thorn elk]

Have you tried expanding the sigma?

$\sum_{r=1}^{n}(r+1).2^r$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[upbeat prism]

I have

and I found that compared to the series in part A

[thorn elk]

The expansion is $2.2+3.2^2+4.2^3+......+(n+1).2^n$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[upbeat prism]

yeah

it's similar to the first series

but i don't know where to go from there

[thorn elk]

Have a look at 1st equation

If we just multiply our 2nd equation by 2, we will obtain the terms of 1st equatiom

equation

[upbeat prism]

wait

let me check

i've been trying to findt he answer for 4 hours

i joined this server just for this question man

i'm gonnna cry if it's that easy

[thorn elk]

You will solve it

Anyways let me help you

[upbeat prism]

wait

[thorn elk]

Alright try

[upbeat prism]

hm

i tried to do it per term

first term checks out

[thorn elk]

(One hint: Subtract 2 from both sides of equation a)

[thorn elk]

No need

[upbeat prism]

what's the purpose of doing this?

[thorn elk]

Did you try this? If yes then

[upbeat prism]

i tried

[thorn elk]

You will see

[thorn elk]

Okay so you get $2.2^2+3.2^3+....+n.2^n=(n-1).2^{n+1}$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[thorn elk]

Multiply this equation by 2

[upbeat prism]

hm

[thorn elk]

And you will find that We have got two IDENTICAL equations

[upbeat prism]

i think

i'm lost

i don't think i'm subtracting 2 from each both sides correctly

[thorn elk]

Do you understand how we got this?

[thorn elk]

Oh

The first term of equation a is 1x2 which is just 2

[upbeat prism]

i'm doing it per term as opposed to

yes

[thorn elk]

And can you see another 2 on far right?

[upbeat prism]

OH

[thorn elk]

So can

you subtract 2 now?

[upbeat prism]

yes

first term for series a is 0

[thorn elk]

So, we get this

Any problems till here?

[upbeat prism]

wait

but why do we only subtract the first term?

shouldn't the subtraction be applied to all the terms since it's a sum?

OH

[thorn elk]

Well, that is the only term which we can cancel out

[upbeat prism]

it's a sum

i got it

[thorn elk]

That's a major misconception

[upbeat prism]

i see

[thorn elk]

Great

[upbeat prism]

oh wow

[thorn elk]

Now lets look at the sigma whose expansion was
The expansion is $2.2+3.2^2+4.2^3+......+(n+1).2^n$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[upbeat prism]

AH

it's similar to the first series

[thorn elk]

I want you to assign the value of $\sum_{r=1}^{n}(r+1).2^r=x$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[thorn elk]

Yes! Just a litlle bit of manipulations

[upbeat prism]

it's the first series without the added 2

ohhhhhhhh

[thorn elk]

Not exactly, the powers of 2 aren't the same

The first series had 2.2^2 but the second one has 2.2

[upbeat prism]

oh yes

OH

so we add 4?

[thorn elk]

No need

I'll show you magic now

[upbeat prism]

😮

[thorn elk]

$$2.2+3.2^2+4.2^3+......+(n+1).2^n=x$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thorn elk]

We have just let that the sum of the series is x

[upbeat prism]

mhm

[thorn elk]

Is that fine?

[upbeat prism]

yes

[thorn elk]

Now, I want you to multiply 2 on both sides

[upbeat prism]

Oh

the power

of 2

increases for each term

[thorn elk]

And you need to know this property $a^m.a^n=a^{m+n}$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[upbeat prism]

yes

because we multiply each term by 2

[thorn elk]

So tell me what will be 2^1.2^2?

[thorn elk]

Yes, we add +1 on powers of 2 on EACH term

[upbeat prism]

OHHHH

which get's us the same series as the first one

minus the 2

[thorn elk]

Almost same

[thorn elk]

Yes

[thorn elk]

We get this

[upbeat prism]

wowzer

[thorn elk]

$2.2^2+3.2^3+....+n.2^n=(n-1).2^{n+1}$

[upbeat prism]

but

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[thorn elk]

Yes?

[upbeat prism]

I got it now

for this question

but i'm still confused as to how you got the answer

like what was your process

because it seems to me

[thorn elk]

Thought process?

[upbeat prism]

kind of

because

everytime i do a series question

which involves comparing two series

the answer only comes from observing "similarities"

or "patterns"

[thorn elk]

Well, pattern observation was done here as well

It was just that it wasn't straightforward

Like we had to subtract 2 and multiply by 2

[upbeat prism]

yeah

[thorn elk]

And then you end up with the same series

But but

We haven't reached our answer

[upbeat prism]

yes

[thorn elk]

Have a look at the last term of this series, its n.2^n

[upbeat prism]

hm

what else is missing?

[thorn elk]

But do you see our series's last term is actually $(n+1).2^{n+1}$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[upbeat prism]

hm

[thorn elk]

So, here we have to apply some brain a

So when the last term of that series was n.2^n our sum was (n-1).2^n+1

[upbeat prism]

wait

how so?

[thorn elk]

This

We have derived this

[upbeat prism]

oh sorry

i thought you meant

• 1

as a different term

[thorn elk]

Oh

I'm just going to write down both the series

[upbeat prism]

ok ok

[thorn elk]

$2.2^2+3.2^3+....+n.2^n=(n-1).2^{n+1}$

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[thorn elk]

$2.2^2+3.2^3+....+n.2^n+(n+1)2^{n+1}=2x$

Where x= sum of the sigma we had

[upbeat prism]

mhm

hm

[thorn elk]

This series has all the terms till n.2^n with an extra

(n+1)2^n+1

[upbeat prism]

wait

could you give me a minute and try to digest this

[oblique flame]

dy/dx= ∏sin(jx).Σj/sin(jx)

[thorn elk]

Yeah

[upbeat prism]

oh

i see

[thorn elk]

In case you are wondering what x is here

[upbeat prism]

but

i got it

but in the context of answering the question

[thorn elk]

Yes?

[upbeat prism]

adding the last term isn't needed right?

for the series

since we only need the equation for the sum of the sigma

[thorn elk]

We will have to add this

[upbeat prism]

oh i see

[thorn elk]

Which will be wrong if we don't add the last term

Now think about this

[thorn elk]

when the last term was n.2^n the entire sum was (n-1).2^n+1

[upbeat prism]

yes

[thorn elk]

Now try to guess what will be the entire sum if the last term is (n+1).2^n+1

Using logic this time

[upbeat prism]

(n-1).2^(n+1) + (n+1).2^(n+1)?

i'm not sure with this though

[thorn elk]

Of course

That's right we just need to add the last term

[upbeat prism]

ohhh

adding that term would give us the complete series

[thorn elk]

yeah

[upbeat prism]

i see

i have to go now

i'll try to further uinderstand this question

you have been a huge help

thankss so much\

[thorn elk]

hey and before you go

[upbeat prism]

i can now sleep in peace

yes?

[thorn elk]

Remember to divide the sum by 2

[thorn elk]

bcz you see

What we got was actually twice the real sum

so we divide it by 2

And yeah, its over with that!

[upbeat prism]

oh yeah

it is

[thorn elk]

Hope you get it

[upbeat prism]

the power has a +1

i got it

actually

omg

THANK YOU

I HAVE TO GO NOW THOUGH

THANKS SO MCUH

[thorn elk]

You're welcome

[upbeat prism]

this definitely won't be the last time you see me here though

[thorn elk]

Try to understand the solution now and make sure to do this without looking at the solution once you understand it!

[upbeat prism]

i have another series question here

i'll try to apply what i've learnned

[thorn elk]

feel free to ask your doubts

[upbeat prism]

thanks again

[thorn elk]

Yeah, try first.

[upbeat prism]

ok

goodbye now

!

[thorn elk]

Bye

[upbeat prism]

.close

[distant dirge]

Unable to parse the channel name

[upbeat prism]

oh

hm

[thorn elk]

You gotta type this to end the conversation-> +close

[upbeat prism]

+close