If $n$ is a positive integer and $0 \le r \le n$, then
$$\begin{array}{rcl}
(a+b)^n & = & C^n_0a^n+C^n_1a^{n-1}b+C^n_2a^{n-2}b^2+\ldots+C^n_ra^{n-r}b^r+\ldots+C^n_nb^n \\
& = & \dsum_{r=0}^nC^n_ra^{n-r}b^r
\end{array}$$
Since $C^n_r=C^n_{n-r}$, the binomial theorem can be rewritten as
$$\begin{array}{rcl}
(a+b)^n & = & C^n_0b^n+C^n_1ab^{n-1}+C^n_2a^2b^{n-2}+\ldots+C^n_ra^rb^{n-r}+\ldots+C^n_na^n \\
& = & \dsum_{r=0}^nC^n_ra^rb^{n-r}
\end{array}$$
$\begin{array}{cl}
& (1-2x)^5 \\
= & C^5_0 (1)^5+C^5_1(1)^4(-2x)+C^5_2(1)^3(-2x)^2+C^5_3(1)^2(-2x)^3+C^5_4(1)(-2x)^4+C^5_5(-2x)^5 \\
= & 1-10x+40x^2-80x^3+80x^4-32x^5
\end{array}$
- Expand $(1+3x)^n$ in ascending powers of $x$ up to the term $x^2$, where $n$ is a positive integer.
- If the coefficient of $x^2$ in the expansion in (a) is $90$, then find the value of $n$.
-
$\begin{array}{cl}
& (1+3x)^n \\
= & C^n_0(1)^n+C^n_1(1)^{n-1}(3x)+C^n_2(1)^{n-2}(3x)^2+\ldots \\
= & 1+3nx+\dfrac{9n(n-1)}{2}x^2+\ldots
\end{array}$ - Since the coefficient of $x^2$ is $90$, by the result of (a), we have
$\begin{array}{rcl}
\dfrac{9n(n-1)}{2} & = & 90 \\
n(n-1) & = & 20 \\
n^2-n-20 & = & 0 \\
(n-5)(n+4) & = & 0
\end{array}$$\therefore n=5$ or $n=-4$ (rejected).
