$a,b\in\mathbb{R}$ ve $n\in\mathbb{N}$ olmak üzere
$(a+b)^n=\sum_{k=0}^n\dbinom{n}{k}a^{n-k}b^k$
olduğunu tümevarım metoduyla gösterelim.
$\boldsymbol{n=1}$ için
$(a+b)^1=a+b=\dbinom{1}{0}a+\dbinom{1}{1}b=\sum_{k=0}^1\dbinom{1}{k}a^{1-k}b^k$
eşitlik doğru.
$\boldsymbol{n=t}$ için
$(a+b)^t=\sum_{k=0}^t\dbinom{t}{k}a^{t-k}b^k$
ifadesinin doğru olduğunu kabul edelim.
$\boldsymbol{n=t+1}$ için
$(a+b)^{t+1}=(a+b)^{t}(a+b)$
$=(a+b)\Bigg[\sum_{k=0}^t\dbinom{t}{k}a^{t-k}b^k\Bigg]$
$=(a+b)\Bigg[\dbinom{t}{0}a^{t}b^0+\dbinom{t}{1}a^{t-1}b^1+\dbinom{t}{2}a^{t-2}b^2+...+\dbinom{t}{t-2}a^{2}b^{t-2}+\dbinom{t}{t-1}a^{1}b^{t-1}+\dbinom{t}{t}a^{0}b^t\Bigg]$
$=\dbinom{t}{0}a^{t+1}b^0+\boldsymbol{\dbinom{t}{1}a^{t}b^1}+\dbinom{t}{2}a^{t-1}b^2+...+\dbinom{t}{t-2}a^{3}b^{t-2}+\dbinom{t}{t-1}a^{2}b^{t-1}+\dbinom{t}{t}a^{1}b^{t}$
$+\boldsymbol{\dbinom{t}{0}a^{t}b^1}+\dbinom{t}{1}a^{t-1}b^2+\dbinom{t}{2}a^{t-2}b^3+...+\dbinom{t}{t-2}a^{2}b^{t-1}+\dbinom{t}{t-1}a^{1}b^{t}+\dbinom{t}{t}a^{0}b^{t+1}$
$=\dbinom{t}{0}a^{t+1}b^0+\boldsymbol{\dbinom{t+1}{1}a^{t}b^1}+\dbinom{t+1}{2}a^{t-1}b^2+...+\dbinom{t+1}{t-1}a^{2}b^{t-1}+\dbinom{t+1}{t}a^{1}b^{t}+\dbinom{t}{t}a^{0}b^{t+1}$
$=\dbinom{t+1}{0}a^{t+1}b^0+\dbinom{t+1}{1}a^{t}b^1+\dbinom{t+1}{2}a^{t-1}b^2+...+\dbinom{t+1}{t-1}a^{2}b^{t-1}+\dbinom{t+1}{t}a^{1}b^{t}+\dbinom{t+1}{t+1}a^{0}b^{t+1}$
$=\sum_{k=0}^{t+1}\dbinom{t+1}{k}a^{t+1-k}b^k$
O halde $\forall n\in\mathbb{N}$ için Binom Teoremi sağlanır.
$\therefore$