Binomial Theorem (1 Viewer)

scardizzle · Sep 27, 2009

Hey guys, having trouble with this question:

In the expansion (1+x+kx^2)^9 in ascending powers of x, the coefficient of x^2 is zero find the value of k.

there are so many coeff of x^2 im sure there is another way then expanding everything out.

Thanks in advance

Drongoski · Sep 27, 2009

If I am not mistaken:

$\100dpi (1 + x + kx^{2})^{9}\ \ =\ \ ([1+x]+kx^{2})^{9}\\ \\ =\ \ \binom{9}{0}[1+x]^{9}(kx^{2})^{0}\ +\ \binom{9}{1}[1+x]^{8}(kx^{2})^{1}\ +\ \cdots \\ \\ \ \ =\ \ \binom{9}{0}[1 + 9x + \binom{9}{2}x^{2}+\cdots ]\times 1\ +\ \binom{9}{1}[1+8x+\cdots ]kx^2 \ +\ \cdots \\ \\ \therefore\ \ term\ in\ x^{2}\ =\ \binom{9}{0}\binom{9}{2}x^{2}\times 1\ +\ \binom{9}{1}\times 1\times kx^{2}\\ \\ \ \ =\ \ 36x^{2}+9kx^{2}\ \ =\ \ (9k+36)x^{2}\ \ =\ \ 0\times x^{2}\ \ \Rightarrow \ \ k = \ -4$

scardizzle · Sep 28, 2009

Thanks, i get it now

Drongoski · Sep 28, 2009

My regular coffee mate suggested today using the multinomial theorem (a generalisation of the binomial - not in the syllabus of course) for this problem. I thought some of you may find it of some interest.

The terms in the multinomial expansion of (1+x+kx²)⁹ containing x² are for 1⁷.x².(kx²)⁰ and for 1⁸.x⁰.(kx²)¹
whose coefficients are respectively:

$\100dpi \frac{9!}{7!\ 2!\ 0!}\ \ =\ \ 36\\ \\ and\ \ \frac{9!}{8!\ 0!\ 1!}\ \ =\ \ 9\ \\ \\ \ \ sum \ of\ coeffs\ of\ x^{2}\ \ = \ \ 36\ +\ 9k \ \ =\ \ 0\ \ \Rightarrow k\ =\ -4$

untouchablecuz · Sep 28, 2009

indeed it is

i have read about it, but i can't seem to find the document for it

edit: here it is

Binomial Theorem (1 Viewer)

scardizzle

Salve!

Drongoski

Well-Known Member

scardizzle

Salve!

Drongoski

Well-Known Member

untouchablecuz

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)