Binomial Theorem (1 Viewer)

taggs-sasuke · Jul 8, 2009

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

Trebla · Jul 9, 2009

One to do it (If you were required to find every term) which is painfully long...
$((1+2x)+3x^2)^5 = \displaystyle\sum_{r=0}^5\binom{5}{r}(1+2x)^r(3x^2)^{5-r}\\\\$But $(1+2x)^r=\displaystyle\sum_{k=0}^r\binom{r}{k}(2x)^k\\\\\Rightarrow((1+2x)+3x^2)^5 = \displaystyle\sum_{r=0}^5\binom{5}{r}\left [\displaystyle\sum_{k=0}^r\binom{r}{k}(2x)^k \right ](3x^2)^{5-r}\\\\=\displaystyle\sum_{r=0}^5\displaystyle\sum_{k=0}^r\binom{5}{r}\binom{r}{k}(2x)^k(3x^2)^{5-r}\\\\=\displaystyle\sum_{r=0}^5\displaystyle\sum_{k=0}^r\binom{5}{r}\binom{r}{k}2^k3^{5-r}x^{k+10-2r}\\\\0 \leq k \leq r \leq 5\\$For constant term we want $ 2r - k = 10\\$Consider $ k = 0, r = 5\\$Term is $\binom{5}{5}\binom{5}{0}2^03^0x^0 = 1\\\\$Note that any higher integer values of $k$ would make $r$ exceed 5$\\\\$For term containing $x$ we want $2r - k = 9\\$For $k = 0, r$ is non-integer. For $k = 2$ or higher, $r$ would exceed 5.$\\ $Consider $ k = 1, r = 5\\$Term is $ \binom{5}{5}\binom{5}{1}2^13^0x^1 = 10x\\\\$

And continue likewise for $x^2$ and $x^3$ ...

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

$((1+2x)+3x^2)^5 = \displaystyle\sum_{r=0}^5\binom{5}{r}(1+2x)^r(3x^2)^{5-r}\\\\$In order to get terms containing $x^3$ or lower degree, the $(3x^2)^{5-r}$ component must have $r=4$ or $5$ because any other value of $r$ will give much higher degrees (e.g. you can immediately inspect that for $r$ = 3 for example, gives $(3x^2)^2 = 9x^4$ and if this is multiplied by any polynomial it will never give something with degree less than 3)$\\\\$Consider $r = 4\\$Term is $\binom{5}{4}(1+2x)^4(3x^2)=15x^2(16x^4+32x^3+24x^2+8x+1)\\$Extracting terms of degree 3 or lower gives: $\\15x^2+120x^3\\\\$Consider $r = 5\\$Term is $\binom{5}{5}(1+2x)^5=32x^5+80x^4+80x^3+40x^2+10x+1\\$Extracting terms of degree 3 or lower gives: $\\80x^3+40x^2+10x+1\\\\$Finally, add those terms up giving the first 4 terms of the full expansion to be: $\\1+10x+55x^2+200x^3$

Pwnage101 · Jul 9, 2009

nice solution

Aquawhite · Jul 9, 2009

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

taggs-sasuke · Jul 9, 2009

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:

$(1+2x+3x^2)^5$

$= [1+(2x+3x^2)]^5$

$= (1+y)^5$
where
$y = 2x+3x^2$

$= 1+5y+10y^2+10y^3+...$

The powers of y higher than $y^3$ are not required as they produce terms in x with powers higher than $x^3$ .

Now $5y = 5(2x+3x^2) = 10x+15x^2$

$10y^2 = 10(2x+3x^2)^2 = 40x^2+120x^3+...$

I don't understand how the below was reached:

Hence

$(1+2x+3x^2)^5 = 1 + 10x + 55x^2 + 200x^3+...$

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

I understand everything before it.

Trebla · Jul 9, 2009

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 · Jul 10, 2009

Trebla said:
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.

Binomial Theorem (1 Viewer)

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