How do you use the binomial series to expand (1+ax)^10?

Jan 10, 2016

$C \left(10 , 0\right) + C \left(10 , 1\right) a + C \left(10 , 2\right) {a}^{2} + C \left(10 , 3\right) {a}^{3} + C \left(10 , 4\right) {a}^{4} + C \left(10 , 5\right) {a}^{5} + C \left(10 , 6\right) {a}^{6} + C \left(10 , 7\right) {a}^{7} + C \left(10 , 8\right) {a}^{8} + C \left(10 , 9\right) {a}^{9} + C \left(10 , 10\right) {a}^{10}$, where C(10,i) is the binomial coefficient representing the number of different ways to choose i items from a set of 10, namely  (10!) / (i! (n-i)!).

Explanation:

In general, for ${\left(x + y\right)}^{k}$, the expansion is $C \left(k , 0\right) {x}^{k} {y}^{0} + C \left(k , 1\right) {x}^{k - 1} {y}^{1} + C \left(k , 2\right) {x}^{k - 2} {y}^{2} + \ldots . C \left(k , k - 1\right) {x}^{1} {y}^{k - 1} + C \left(k , k\right) {x}^{0} {y}^{k}$.

Note that, for example, $C \left(4 , 0\right) = C \left(4 , 4\right) = 1 , C \left(4 , 1\right) = C \left(4 , 3\right) = 4 , C \left(4 , 2\right) = 6$
and Pascal's triangle gives a diagram representing binomial coefficients, namely:

$$                                               1
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


etc.
The sequence of numbers in the i-th row are, respectively, $C \left(1 , 0\right) , C \left(i , 1\right) , C \left(i , 2\right) , C \left(i , 3\right) ,$ etc.

The number $C \left(i , j\right)$ is the sum of the two numbers $C \left(i - 1 , j - 1\right)$ and $C \left(i - 1 , j\right)$ just above it in the diagram and to the left and right, respectively.