# How do you use the binomial series to expand  (2 + x)^11 ?

Dec 6, 2015

Multiply a row of Pascal's triangle by a sequence of descending powers of $2$ to find:

${\left(2 + x\right)}^{11} = 2048 + 11264 x + 28160 {x}^{2} + 42240 {x}^{3} + 42240 {x}^{4} + 29548 {x}^{5} + 14784 {x}^{6} + 5280 {x}^{7} + 1320 {x}^{8} + 220 {x}^{9} + 22 {x}^{10} + {x}^{11}$

#### Explanation:

${\left(2 + x\right)}^{11} = {\sum}_{k = 0}^{11} \left(\begin{matrix}11 \\ k\end{matrix}\right) {2}^{11 - k} {x}^{k}$

We can find the values of $\left(\begin{matrix}11 \\ k\end{matrix}\right)$ from Pascal's triangle:

Write out the row beginning $1$, $11$:

$1 , 11 , 55 , 165 , 330 , 462 , 462 , 330 , 165 , 55 , 11 , 1$

Write out the powers of $2$ in descending order from ${2}^{11}$ to ${2}^{0}$:

$2048 , 1024 , 512 , 256 , 128 , 64 , 32 , 16 , 8 , 4 , 2 , 1$

Multiply these two sequences together to get:

$2048 , 11264 , 28160 , 42240 , 42240 , 29548 , 14784 , 5280 , 1320 , 220 , 22 , 1$

These are the coefficients to find:

${\left(2 + x\right)}^{11} = 2048 + 11264 x + 28160 {x}^{2} + 42240 {x}^{3} + 42240 {x}^{4} + 29548 {x}^{5} + 14784 {x}^{6} + 5280 {x}^{7} + 1320 {x}^{8} + 220 {x}^{9} + 22 {x}^{10} + {x}^{11}$