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Binomial theorem

Jun 12, 2023

# Use Mathematical Induction to prove

  1. $$(a+b)^n = a^n + \dbinom{n}{1}a^{n-1}b +
    \dbinom{n}{2}a^{n-2}b^2+\dots+\dbinom{n}{r}a^{n-r}b^r+\dots+ \dbinom{n}{n-1}ab^{n-1} + b^n$$
    is trivially true for n = 1, 2
  2. Let k > 2 $$\begin{align*}
    &(a+b)^{k+1}\
    &= (a+b)(a+b)^k\
    &= (a+b)\Bigg(a^k + \dbinom{k}{1}a^{k-1}b + \dbinom{k}{2}a^{k-2}b^2+\dots
    +\dbinom{k}{r}a^{k-r}b^r +\dots+ \dbinom{k}{k-1}ab^{k-1} + b^k \Bigg)\
    &\begin{aligned}[t]
    &= a^{k+1} + \Bigg[1+\dbinom{k}{1}\Bigg]a^kb + \Bigg[\dbinom{k}{1}+\dbinom{k}{2}\Bigg]a^{k-1}b^2 + \dotsb\
    &\dotsb + \Bigg[\dbinom{k}{r-1} + \dbinom{k}{r}\Bigg]a^{k-r + 1}b^r + \dotsb + \Bigg[\dbinom{k}{k-1} + 1\Bigg]ab^{k} + b^{k+1}.
    \end{aligned}
    \end{align*}$$
  3. Use $\dbinom{n}{r-1} + \dbinom{n}{r} = \dbinom{n+1}{r}, \qquad\text{for}\quad 0 < r \leq n$
  4. QED.