Question

Solve 59 question

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Answer 1

Note: your "N: = my "n" here.

I can't believe I'm doing this,

Here wo go fellas,

Lemme just write down what the question is providing and asking,
For the sake of people like me who didn't get it at the first glance,

We've been given,
$y = {x}^{4} {e}^{2x}$
And

gotta find,
$\frac{ {d}^{10}( {x}^{4} {e}^{2x}) }{d {x}^{10} } \: (0) = n \times {2}^{10}$

Sounds easy solly easier now?
No?

Bear with me then,

You can use Leibniz Theorem for product rule or do direct differentiation step wise, depends on how lazy your as is.

So let's say you have a diffrential term as,
uv (cause this gives product vibes)

So by libnitz formula you're gonna get the first differentiation as ,

${(uv)}^{'} = uv' + vu'$
The second as,

$(uv' + u'v)' = (uv')' + (u'v)' \\ = (uv'' + u'v') + (u''v + u'v') \\ = uv'' + 2u'v' + u''v'$

Continuing this to third step gives,
$uv''' + 3u'v'' + 3u''v' + u'''v$

Doing it for few more steps it gives back a patter you might have noticed in binomial theorem chapter (Pascal's traingle).

So in the end the gerneral form we get is in the attachment.

Now,
I prefer the differentiation step wise method cause it makes things look oh so easy,
So imma use that

The first differentiation will give,
$4 {x}^{3} {e}^{2x} + {e}^{2x} {2}x^{4}$

The second will give,

$4 {e}^{2x} {x}^{4} + 16 {e}^{2x} {x}^{3} + 12 {e}^{2x} {x}^{2}$

The third will,
$8 {e}^{2x} {x}^{4} + 48 {e}^{2x} {x}^{3} + 72 {e}^{2x} {x}^{2} + 24 {e}^{2x} x$

and this will keep going till 10th step where we will get the result as,

$1024 {e}^{2x} {x}^{4} + 20480 {e}^{2x} {x}^{3} + 138240 {e}^{2x} {x}^{2} + 36864 {e}^{2x} x + 322560 {e}^{2x}$

Put x =0 and you'll have,

(all other terms as 0 but not this one)

$322560 {e}^{2(0)} = n \times {2}^{10} \\ 322560 = n \times {2}^{10} \\315 \times {2}^{10} = n \times {2}^{10} \\ \\ n = 315$

Add the terms and tada, We have the required sum as 9.

(And oh thank you for telling me that n%2=0).

Hope you're not annoyed as should've been.