(1+x)^8

To expand the expression \((1+x)^8\), we can use the binomial theorem. The binomial theorem states that for any positive integer \(n\) and any real numbers \(a\) and \(b\), \((a+b)^n\) can be expanded as:

\[(a+b)^n = \binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2} + \ldots + \binom{n}{n-1}a^{1}b^{n-1} + \binom{n}{n}a^{0}b^{n}\]

In our case, \(a=1\) and \(b=x\), and \(n=8\). So, we have:

\((1+x)^8 = \binom{8}{0}1^{8}x^{0} + \binom{8}{1}1^{7}x^{1} + \binom{8}{2}1^{6}x^{2} + \ldots + \binom{8}{7}1^{1}x^{7} + \binom{8}{8}1^{0}x^{8}\)

Solving each term:

\[\binom{8}{0}1^{8}x^{0} = 1\]

\[\binom{8}{1}1^{7}x^{1} = 8x\]

\[\binom{8}{2}1^{6}x^{2} = 28x^{2}\]

\[\binom{8}{3}1^{5}x^{3} = 56x^{3}\]

\[\binom{8}{4}1^{4}x^{4} = 70x^{4}\]

\[\binom{8}{5}1^{3}x^{5} = 56x^{5}\]

\[\binom{8}{6}1^{2}x^{6} = 28x^{6}\]

\[\binom{8}{7}1^{1}x^{7} = 8x^{7}\]

\[\binom{8}{8}1^{0}x^{8} = x^{8}\]

Putting all the terms together, we get:

\((1+x)^8 = 1 + 8x + 28x^{2} + 56x^{3} + 70x^{4} + 56x^{5} + 28x^{6} + 8x^{7} + x^{8}\)

Therefore, \((1+x)^8\) expands to \(1 + 8x + 28x^{2} + 56x^{3} + 70x^{4} + 56x^{5} + 28x^{6} + 8x^{7} + x^{8}\).