Binomial Theorem

[IloveManUtd]

In my question

Find, in ascending powers of x, the first 3 terms in the expansion of (2+3x-4x[sup:2czwrjur]2[/sup:2czwrjur])[sup:2czwrjur]5[/sup:2czwrjur]

Well I arranged it into [(2+3x)[sup:2czwrjur]5[/sup:2czwrjur] +(-4x[sup:2czwrjur]2[/sup:2czwrjur])[sup:2czwrjur]5[/sup:2czwrjur]]. Then I'm stuck. Please help.

 

[Deleted member 4993]

In my question

Find, in ascending powers of x, the first 3 terms in the expansion of (2+3x-4x[sup:12c2uqls]2[/sup:12c2uqls])[sup:12c2uqls]5[/sup:12c2uqls]

Well I arranged it into [(2+3x)[sup:12c2uqls]5[/sup:12c2uqls] +(-4x[sup:12c2uqls]2[/sup:12c2uqls])[sup:12c2uqls]5[/sup:12c2uqls]]. <<< That is not correct Then I'm stuck. Please help.

Please search your textbook/classnotes/google and find expression for (a + b)[sup:12c2uqls]5[/sup:12c2uqls]

Post what you found - we'll work from there.

 

[galactus]

Have you heard of Pascal's triangle for the binomial expansion?.

Well, this is Pascal's tetrahedron for the trinomial expansion.

$\displaystyle (x+y+z)^{5}=\binom{5}{0}z^{5}y^{0}x^{0}+\binom{5}{1}z^{4}y^{1}x^{0}+\binom{5}{2}z^{3}y^{2}x^{0}+.....$

See the pattern?. Sub in your x, y, and z for a start.

 

[Deleted member 4993]

This is an old problem from:

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
International General Certificate of Secondary Education

 

[mmm4444bot]

This is an old problem from:

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

You took that exam in 1904, yes?

 

[Deleted member 4993]

Re:

This is an old problem from:

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

You took that exam in 1904, yes?

Actually the most recent test with that problem came in 2008.

But yes - I did take that test in 1904 (well that's not true - my grandfather was born in 1904).... at least feels like that. Time flies like an arrow (and fruitflies like a rotten banana)....

 

[soroban]

Hello, IloveManUtd!

Find, in ascending powers of $\displaystyle x$, the first 3 terms in the expansion of: .$\displaystyle (2+3x-4x^2)^5$

galactus mentioned Pascal's Tetrahedron for the expansion of trinomials.


$\displaystyle \text{For }(a + b + c)^5\text{, the terms look like this:}$

. . $\displaystyle \begin{array}{cccccccccccc}&&&&& a^5 \\ \\ &&&& 5a^4b && 5a^4c \\ \\ &&& 10a^3b^2 && 20a^3bc && 10a^3c^2 \\ \\ && 10a^2b^3 && 30a^2b^2c && 30a^2bc^2 && 10a^2c^3 \\ \\ & 5ab^4 && 20ab^3c && 20ab^2c^2 && 20 abc^3 && 5 ac^4 \\ \\ b^5 && 5b^4c && 10 b^3c^2 && 10b^2c^3 && 5bc^4 && c^5 \end{array}$&&



$\displaystyle \text{We are given: }\;\begin{Bmatrix}a &=& 2 \\ b &=& 3x \\ c &=& \text{-}4x^2 \end{Bmatrix}$


$\displaystyle \text{Substitute into the Triangle and find the constant term, the coefficient of the }x\text{-term}$
. . $\displaystyle \text{and the coefficient of the }x^2\text{-term.}$


Looking ahead, we see that we need only the first four terms of the Triangle.

. . $\displaystyle a^5 + 5a^4b + 5a^4c + 10a^3b^2$

. . . . . $\displaystyle =\; (2^5) + 5(2^4)(3x) + 5(2^4)(\text{-}4x) + 10(2^3)(3x)^2$

. . . . . $\displaystyle =\;32 + 240x - 320x^2 + 720x^2$

. . . . . $\displaystyle =\;32 + 240x + 400x^2$