# Binomial Expansions

• February 24th 2010, 04:47 AM
teddybear67
Binomial Expansions
Q1)Find, in its simplest form, the coefficient of $x^r$ in the expansion, in ascending powers of $x$ , of $\frac{1}{x-3}$

Q2) Expand $(1+y)^14$ as a series of ascending powers of $y$ up to and including the term in $y^3$, Simplify the coefficients.

I got the above part right and derived at the answer of $1 + 14y + 91y^2 + 364y^3$

However, its the following part that i dont get

In The expansion of $(1 + x + kx^2)^14$, where $k$ is a constant, the coefficient of $x^3$ is zero. By writing $x + kx^2$ as $y$, or otherwise, find the value of k.

Any help is appreciated ^^
• February 24th 2010, 05:31 AM
Quote:

Originally Posted by teddybear67
Q1)Find, in its simplest form, the coefficient of $x^r$ in the expansion, in ascending powers of $x$ , of $\frac{1}{x-3}$

Q2) Expand $(1+y)^14$ as a series of ascending powers of $y$ up to and including the term in $y^3$, Simplify the coefficients.

I got the above part right and derived at the answer of $1 + 14y + 91y^2 + 364y^3$

However, its the following part that i dont get

In The expansion of $(1 + x + kx^2)^14$, where $k$ is a constant, the coefficient of $x^3$ is zero. By writing $x + kx^2$ as $y$, or otherwise, find the value of k.

Any help is appreciated ^^

hi

begin with some substitution as hinted by the question

1+14(x+kx^2)+91(x+kx^2)^2+364(x+kx^2)^3+91(x^2+2kx ^3+k^2x^4)+364(x^3+3kx^4+3x(kx^2)^2+(kx^2)^3)

Given that the coefficient of $x^3$ is 0 , so we need to know the coefficient of x^3 only . To expand the whole thing would be tedious , so we expand those parts where $x^3$ is .

$..... 91x^2+182kx^3 + ... +364x^3+....$

coefficient of x^3 : $182k+364\Rightarrow k=-2$
• February 24th 2010, 01:08 PM
teddybear67
Thanks!
But do ya know how to do the top one too? :/
• February 25th 2010, 02:28 AM
HallsofIvy
$\frac{1}{x- 3}= (x- 3)^{-1}$

Use the "generalized binomial theorem"
$(a+ b)^r= \sum_{i=0}^\infty\begin{pmatrix}r \\ i\end{pmatrix}a^ib^{r-i}$
Where, for r not a positive integer,
$\begin{pmatrix}r \\ i\end{pmatrix}$ $= \frac{r(r-1)\cdot\cdot\cdot(r- i+1)}{i!}$
With r not a positive integer, r- i+ 1 is never equal to 0 so this is an infinite series.