Binomial Theorem

Pascals Theorem This is using pascal triangle to expand binomials. So a degree 3 binomial will match the coefficients with that row in the pascal’s triangle.

Shorthand notation

$(x+y{)}^n=\sum _{i-0}^n{=}_n{C}_i{x}^{i-1}{y}^i$

Example

$(x+y{)}^4$ 4th row of pascal triangle: 1 4 6 4 1 $(x+y{)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$

Example 2

$(1+bx{)}^n=1-3x+\frac{15}{4}{x}^2$

term 1:

${}_n{C}_0(1{)}^n(bx{)}^0=1$

term 2:

${}_n{C}_1(1{)}^{n-1}(bx{)}^1=-3x$ ${}_n{C}_1=\frac{n!}{n-1(1!)}=n$ $nbx=-3x$ $nb=-3$ $b=\frac{-3}{n}$

term 3:

${}_n{C}_2(1{)}^{n-2}(bx{)}^2=\frac{15}{4}{x}^2$ $=\frac{n!}{(n-2)!2!}{b}^2{x}^2=\frac{15}{4}{x}^2$ $=\frac{(n)(n-1)}{2}{b}^2{x}^2=\frac{15}{4}{x}^2$ $\frac{(n)(n-1)}{2}{b}^2=\frac{15}{4}$ $\frac{(n)(n-1)}{2}{\left(\frac{3}{n}\right)}^2=\frac{15}{4}$ $\frac{(n)(n-1)}{2}\left(\frac{9}{n^2}\right)=\frac{15}{4}$ $\frac{(9)(n-1)}{2n}=\frac{15}{4}$ $3b(n-1)=30n$ $3bn-3b=30n$ $6n=3b$ $n=6$ $b=-1/2$