Proof by Differentiating Binomial Expansions

Part 1

Show that for all positive integers $n$,

$x \left[ (1+x)x^{n-1} + (1+x)^{n-2} + \cdots + (1+x)^2 + (1+x) +1 \right] = (1+x)^n-1$.

\require{AMSsymbols} \begin{align} \displaystyle \text{LHS} &= x \left[(1+x)x^{n-1} + (1+x)^{n-2} + \cdots + (1+x)^2 + (1+x) +1 \right] \\ &= x \times \frac{1\left[ (1+x)^n-1 \right]}{(1+x)-1} \ \ \ \color{green}{\text{first term}=1, \text{common ratio}=1+x, \text{terms}=n} \\ &= x \times \frac{(1+x)^n-1 }{x} \\ &= (1+x)^n-1 \\ &= \text{RHS} \end{align}

Part 2

Hence show that for $1 \le k \le n$,

$\begin{pmatrix} n-1 \\ k-1 \end{pmatrix} + \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix}$.

\require{AMSsymbols} \begin{align} (1+x)^n-1 &= \begin{pmatrix} n \\ 0 \end{pmatrix} + \begin{pmatrix} n \\ 1 \end{pmatrix}x + \begin{pmatrix} n \\ 2 \end{pmatrix}x^2 + \cdots + \begin{pmatrix} n \\ k \end{pmatrix} + \cdots + \begin{pmatrix} n \\ n \end{pmatrix}x^n – 1 \end{align}
Therefore the coefficient of $x^k$ is $\begin{pmatrix} n \\ k \end{pmatrix}$.
\require{AMSsymbols} \begin{align} \text{LHS} &= x(1+x)^{n-1} + x(1+x)^{n-2} + x(1+x)^{n-3} + \cdots + x(1+x)^{k-1} + \cdots \\ &= \begin{pmatrix} n-1 \\ 0 \end{pmatrix}x + \begin{pmatrix} n-1 \\ 1 \end{pmatrix}x^2 + \cdots + \begin{pmatrix} n-1 \\ k-1 \end{pmatrix}x^k + \cdots \\ &+ \begin{pmatrix} n-2 \\ 0 \end{pmatrix}x + \begin{pmatrix} n-2 \\ 1 \end{pmatrix}x^2 + \cdots + \begin{pmatrix} n-2 \\ k-1 \end{pmatrix}x^k + \cdots \\ &\cdots \\ &+ \begin{pmatrix} k-1 \\ 0 \end{pmatrix}x + \begin{pmatrix} k-1 \\ 1 \end{pmatrix}x^2 + \cdots + \begin{pmatrix} k-1 \\ k-1 \end{pmatrix}x^k + \cdots \\ & \cdots \end{align}
The coefficient of $x^k$ is $\begin{pmatrix} n-1 \\ k-1 \end{pmatrix} + \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + \cdots + \begin{pmatrix} k-1 \\ k-1 \end{pmatrix}$
Equating coefficients
$\therefore \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} + \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + \cdots + \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix}$

Part 3

Show that $n \begin{pmatrix} n-1 \\ k \end{pmatrix} = (k+1) \begin{pmatrix} n \\ k+1 \end{pmatrix}$.

\displaystyle \begin{align} n \begin{pmatrix} n-1 \\ k \end{pmatrix} &= \frac{n(n-1)!}{k!(n-1-k)!} \\ &= \frac{n(n-1)(n-2)(n-3) \cdots 1}{k![n-(k+1)]!} \\ &= \frac{n!}{k![n-(k+1)]!} \times \frac{k+1}{k+1} \\ &= \frac{n!(k+1)}{(k+1)![n-(k+1)]!} \\ &= (k+1) \times \frac{n!}{(k+1)![n-(k+1)]!} \\ &= \begin{pmatrix} n \\ k+1 \end{pmatrix} \end{align}

Part 4

Show the following for $1 \le k \lt n$.
$(n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} = k \begin{pmatrix} n \\ k+1 \end{pmatrix}$

\require{AMSsymbols} \begin{align} (1+x)^n -1 &= x \left[ (1+x)x^{n-1} + (1+x)^{n-2} + \cdots + (1+x)^2 + (1+x) +1 \right] \color{green}{\cdots \text{Part 1}} \\ \frac{d}{dx} \left[(1+x)^n-1 \right] &= n(1+x)^{n-1} \\ &= n \left[ \begin{pmatrix} n-1 \\ 0 \end{pmatrix}x^0 + \begin{pmatrix} n-1 \\ 1 \end{pmatrix}x + \begin{pmatrix} n-1 \\ 2 \end{pmatrix}x^2 + \cdots + \begin{pmatrix} n-1 \\ k \end{pmatrix}x^k + \cdots \right] \\ \text{The coefficient of } x^k &= n \begin{pmatrix} n-1 \\ k \end{pmatrix} \\ n \begin{pmatrix} n-1 \\ k \end{pmatrix} &= (k+1) \begin{pmatrix} n \\ k+1 \end{pmatrix} \color{green}{\cdots \text{Part 3}} \\ \frac{d}{dx} &x \left[ (1+x)x^{n-1} + (1+x)^{n-2} + \cdots + (1+x)^2 + (1+x) +1 \right] \\ &= x \left[ (n-1)(1+x)^{n-2} + (n-2)(1+x)^{n-3} + \cdots + 2(1+x) + 1 \right] \\ &+ (1+x)x^{n-1} + (1+x)^{n-2} + \cdots + (1+x)^2 + (1+x) +1 \\ &= \left[ (n-1)x(1+x)^{n-2} + (n-2)x(1+x)^{n-3} + \cdots + kx(1+x)^{k-1} + \cdots + 2x(1+x)+x \right] \color{red}{\cdots (1)} \\ &+ (1+x)^{n-1} + (1+x)^{n-2} + \cdots + (1+x) + 1 \color{red}{\cdots (2)} \\ \text{The coefficient of } x^k \text{ for } &(n-1)x(1+x)^{n-2} \text{ is } (n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} \\ \text{The coefficient of } x^k \text{ for } &(n-2)x(1+x)^{n-3} \text{ is } (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} \\ \text{The coefficient of } x^k \text{ for } &kx(1+x)^{k-1} \text{ is } k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} \\ \text{The coefficient of } x^k \text{ in } &\color{red}{(1)} \text{ is } (n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} \\ \text{The coeffifient of } x^k \text{ in } &\color{red}{(2)} \text{ is } \begin{pmatrix} n-1 \\ k \end{pmatrix} + \begin{pmatrix} n-2 \\ k \end{pmatrix} + \cdots + \begin{pmatrix} k \\ k \end{pmatrix} \\ \text{The coefficient of } x^k \text{ is } &(n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} \\ &+ \left[ \begin{pmatrix} n-1 \\ k \end{pmatrix} + \begin{pmatrix} n-2 \\ k \end{pmatrix} + \cdots + \begin{pmatrix} k \\ k \end{pmatrix} \right] \\ &= (k+1) \begin{pmatrix} n \\ k+1 \end{pmatrix} \end{align} \\ \text{Using the result of Part 2} \begin{align} (n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} + \begin{pmatrix} n \\ k+1 \end{pmatrix} &= (k+1) \begin{pmatrix} n \\ k+1 \end{pmatrix} \\ (n-1) \begin{pmatrix} n-2 \\ k-1 \end{pmatrix} + (n-2) \begin{pmatrix} n-3 \\ k-1 \end{pmatrix} + \cdots + k \begin{pmatrix} k-1 \\ k-1 \end{pmatrix} &= (k+1) \begin{pmatrix} n \\ k+1 \end{pmatrix}-\begin{pmatrix} n \\ k+1 \end{pmatrix} \\ &= \begin{pmatrix} n \\ k+1 \end{pmatrix}(k+1-1) \\ &= k \begin{pmatrix} n \\ k+1 \end{pmatrix} \end{align}

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