Question

How do you use the binomial series to expand `(x + 3)^12`?

Answer 1

`(x+3)^12`

`=sum_(k=0)^12 ((12),(k)) 3^k x^(12-k)`

`=x^12+36x^11+594x^10+5940x^9+40095x^8+192456x^7+673596x^6+1732104x^5+3247695x^4+4330260x^3+3897234x^2+2125764x+531441`

Explanation:

You can use Pascal's triangle to obtain the values of `((12),(0))`, `((12),(1))`,...,`((12),(12))`

Write out the row starting `1`, `12`,...

`1`, `12`, `66`, `220`, `495`, `792`, `924`, `792`, `495`, `220`, `66`, `12`, `1`

Write out ascending powers of `3` from `3^0` to `3^12`...

`1`, `3`, `9`, `27`, `81`, `243`, `729`, `2187`, `6561`, `19683`, `59049`, `177147`, `531441`

Multiply both sequences together to get:

`1`, `36`, `594`, `5940`, `40095`, `192456`, `673596`, `1732104`, `3247695`, `4330260`, `3897234`, `2125764`, `531441`

These are the coefficients we need since:

`(x+3)^12 = sum_(k=0)^12 ((12),(k)) 3^k x^(12-k)`

So:

`(x+3)^12`

`=x^12+36x^11+594x^10+5940x^9+40095x^8+192456x^7+673596x^6+1732104x^5+3247695x^4+4330260x^3+3897234x^2+2125764x+531441`