Series
\[y_1(x)=4+\frac{4}{8}x^2+\frac{4}{4\cdot 4\cdot 3}x^4+\frac{24 \cdot 4}{4\cdot 4\cdot 3\cdot 5\cdot6}x^6\] i have to write the sigma notation of ^the series
Tagging doesnt work in the question... Remind me to fix that
they are not online so doesn't matter lol
Yea but if the haven't disabled notifications they would receive a email lol
@Vocaloid Help Help Help pleaseeeee D:
I'm not sure, I believe if you start n at 0 you can write the x part as x^2n but I'm uncertain about the coefficients @hero
yeah i'm also confused about the numbers.
\[=\large\rm \frac{4}{1!}+\frac{4}{4\cdot 2!} x^2+\frac{4}{4 \cdot 2 \cdot 3!}x^4+\frac{ 24 \cdot4 }{ 4!\cdot 2\cdot 6\cdot 5}x^6\]
Thanks for the update. The original post was not making any sense at all.
It's same as the top post.
The top post does not have factorials
We can't write this as sigma notation, so good thing i don't have to simplify anymore..
yeah I've simplified to see the pattern but didn't work out.
What did you simplify it to?
the 2nd one.
You can simplify it even further than that
do you mean reducing fractions? yeah but i meant to say that i don't have to see the pattern. I just need the terms as final solution.
Yes, reducing the fractions.
i'll definitely do that :=))
Post your result
result ?
Yes, post the expression you get after reducing the fractions.
lol ok sir
\[y(x)= 4+2x^2+12x^4+\frac{2}{90}x^6\]
I got completely different coefficients
sorry 1/45
Even still
show me your work.
\(\dfrac{4}{4 \cdot 2!} = \dfrac{4}{4 \cdot2\cdot1} = \dfrac{4}{4} \times \dfrac{1}{2 \cdot 1}\)
For example
\[\frac{ 2\cancel{4} \cdot \cancel{ 4 }}{ \cancel{4} \cdot cancel{4} \cdot 3\cdot 5\cdot 6 }\]
Why'd you skip the first one?
\[\frac{ 2\cancel{4} \cdot \cancel{4} }{ \cancel{4} \cdot \cancel{4} \cdot 3 \cdot 5 \cdot 6 }\]
bec you're going to do it for me. thank you
Yep, and you can still simplify even further.
You should get fractional coefficients for all but the first term.
\[\frac{ 2\cancel{4 }\cdot \cancel{4} }{\cancel{ 4} \cdot \cancel{4} \cdot 3 \cdot 5 \cdot 6 }= \frac{2}{90}=\frac{1}{45}\]
Which term is that? You have to start from the beginning. I don't know which term that is.
last one
i start from right
\[\frac{ 2\cancel{4 }\cdot \cancel{4} }{\cancel{ 4} \cdot \cancel{4} \cdot 3 \cdot 5 \cdot \color{Red}{\cancel{6}} }x^\color{Red}{\cancel{6}}= =\frac{2}{15}\]
found a mistake sorry
Of course you did.
I don't feel like typing it all out to be honest.
same
just give me the answer
Hang on

that's wrong ?? :o
Where's the SIX in the numerator. Remember, originally, you had 24 * 4 in the numerator, right?
yes 24 since 4 and 4 are the same so 4 cancels out
And how are you cancelling the \(x^6\) term?
Your work as currently constructed makes no sense.
since 6 and 6 are alike, 6 cancels out
You're cancelling exponents with coefficients. That doesn't make any sense.
ohh i can't do that ??? hmm sorry
NO, you definitely cannot do that.
ohh okay don't yell at me still learning
As far as I know, that is correct.
Unless your 24 isn't 24.
Is 24 supposed to be TWENTY-FOUR? Or is it supposed to be 2 times 4?
Is 24 supposed to be TWENTY-FOUR? Or is it supposed to be 2 times 4?
yeah 24 is 24 or it can 42 right since ab=ba
Because you have it written as TWENTY-FOUR not 2 times 4. Do you understand what I'm saying?
yes sir
If I interpret what you have written literally. \(24\) as you have written \(/ne\ 2 \times 4\) Do you understand?
I'm gonna give up if you don't comprehend that you wrote 24 and not \(2 \times 4\)
it's too wenty for
i have to go thanks hero!!
smh
Good luck. You seem to confuse numbers at certain points. You have to work on that.
lol
i can't believe what i ended up doing
@Nnesha perhaps try posting the problem from the original source itself.

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