# Tổng chuổi số Pascal

## Thí dụ

 ${\displaystyle (x+1)^{1}=}$ ${\displaystyle 1x+1}$ ${\displaystyle (x+1)^{2}=}$ ${\displaystyle 1x^{2}+2x+1}$ ${\displaystyle (x+1)^{3}=}$ ${\displaystyle 1x^{3}+3x^{2}+3x+1}$ ${\displaystyle (x+1)^{4}=}$ ${\displaystyle 1x^{4}+4x^{3}+6x^{2}+4x+1}$ ${\displaystyle (x+1)^{5}=}$ ${\displaystyle 1x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1}$

## Tam giác Pascal

Tam giác Pascal cho biet so dang truoc bien so luy thua

                                     1     1
1     2     1
1     3     3     1
1     4     6     4     1
1     5     10    10    5     1
1     6     15    20    15    6     1
1     7     21    35    35    21    7     1
1     8     28    56    70    56    28    8     1
1     9     36    84    126   126   84    36    9     1
1     10    45    120   210   252   210   120   45    10    1
1      11    55    165   330   462   462   330   165   55   11     1

${\displaystyle (a+b)^{o}=1}$
${\displaystyle (a+b)^{1}=a+b}$
${\displaystyle (a+b)^{2}=(a+b)(a+b)=a^{2}+ab+ab+b^{2}=a^{2}+2ab+b^{2}}$

## Công thức tổng chuổi số Pascal

• ${\displaystyle (x+y)^{n}=\sum _{r=0}^{n}{n \choose r}x^{r}y^{n-r}}$
${\displaystyle (x+y)^{n}={n \choose 0}x^{0}y^{n}+{n \choose 1}x^{1}y^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+{n \choose {n-1}}x^{n-1}y^{1}+{n \choose n}x^{n}y^{0}}$
${\displaystyle (x+y)^{n}=y^{n}+nxy^{n-1}+{n \choose 2}x^{2}y^{n-2}+\dots +{n \choose {n-2}}x^{n-2}y^{2}+nx^{n-1}y+x^{n}}$

Với

${\displaystyle {n \choose r}={\frac {n!}{r!(n-r)!}}}$