دستور ثنائي الحد

دستور ثنائي الحد
( E: Binomial Theorem )
(T: Binom Açılımı )

$(a+b)^n=\sum^n_{k=0} {n\choose k} a^{n-k}b^k~,~n\in \Bbb N$
${n\choose k}=\frac {n(n-1)(n-2)\cdots (n-k+1)}{k(k-1)\cdots 2\cdot 1}$

مثال1:
$(a+b)^3=\sum^3_{k=0} {3\choose k} a^{3-k}b^k={3\choose 0} a^3b^0$
$+{3\choose 1} a^2b^1+{3\choose 2} a^1b^2+{3\choose 3} a^0b^3$
$=a^3+3a^2b+\frac{3\cdot 2}{2}ab^2+b^3$
$=a^3+3a^2b+3ab^2+b^3$
مثال2:
$(a+b)^5=\sum^5_{k=0} {5\choose k} a^{5-k}b^k={5\choose 0} a^5b^0$
$+{5\choose 1} a^4b^1+{5\choose 2} a^3b^2+{5\choose 3} a^2b^3$
$+{5\choose 4} a^1b^4+{5\choose 5} a^5b^0$
$=a^5+5a^4b+\frac{5\cdot 4}{2}a^3b^2+\frac{5\cdot 4\cdot 3}{3\cdot 2}a^2b^3$
$+\frac{5\cdot 4\cdot 3\cdot 2}{4\cdot 3\cdot 2}ab^4+a^5$
$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+a^5$
مثال3:
$(2x+1)^4=\sum^4_{k=0} {4\choose k} (2x)^{4-k}1^k=\sum^4_{k=0} {4\choose k} (2x)^{4-k}$
$={4\choose 0} (2x)^4+{4\choose 1} (2x)^3+{4\choose 2} (2x)^2$
$+{4\choose 3} (2x)^1+{4\choose 4} (2x)^0$
$= 16x^4+32x^3+24x^2+8x+1$