Binomial Theorem (1 Viewer)

taggs-sasuke

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Can you please help me with this question?

I'm sorry if it's really easy but I'm in the process of learning this topic and I can't do it.

Find the expansion of [maths](1+2x+3x^2)^5[/maths] in ascending powers of x up to [maths]x^3[/maths].

Thanks :)
 
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Trebla

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One to do it (If you were required to find every term) which is painfully long...


And continue likewise for and ...

Since the given question doesn't require every term, we can just do:

 

Aquawhite

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Really long big mess... As you may tell, I dislike the binomial stuff.
 

taggs-sasuke

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Thanks for your reply, Trebla, but I haven't learnt that notation yet...
(I'm so sorry for wasting your time.)


This is the solution:






where




The powers of y higher than are not required as they produce terms in x with powers higher than .

Now



I don't understand how the below was reached:

Hence




Can you please explain it to me? (Thank you!) :)

I understand everything before it.
 

Trebla

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It's basically, evaluating the 1 + 5y + 10y² + 10y³ in terms of x:
1 + 5y + 10y² + 10y³ = 1 + 5(2x + 3x²) + 10(2x + 3x²)² + 10(2x + 3x²)³
Expand and collect all like terms with degrees of 3 or less. It's somewhat similar to the method I typed up, it's just that I chose a different pair and didn't bother using y = ....
 

taggs-sasuke

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It's basically, evaluating the 1 + 5y + 10y² + 10y³ in terms of x:
1 + 5y + 10y² + 10y³ = 1 + 5(2x + 3x²) + 10(2x + 3x²)² + 10(2x + 3x²)³
Expand and collect all like terms with degrees of 3 or less. It's somewhat similar to the method I typed up, it's just that I chose a different pair and didn't bother using y = ....
Oh, I get it.

I overlooked something simple again. :(
 

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