(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 \(n\), we have:

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

In this case, \(a = 1\) and \(b = x\), and \(n = 8\). Substituting these values into the formula, we get:

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

Now, we can simplify this expression by calculating the binomial coefficients:

\(\binom{8}{0}\) = 1
\(\binom{8}{1}\) = 8
\(\binom{8}{2}\) = 28
\(\binom{8}{3}\) = 56
\(\binom{8}{4}\) = 70
\(\binom{8}{5}\) = 56
\(\binom{8}{6}\) = 28
\(\binom{8}{7}\) = 8
\(\binom{8}{8}\) = 1

Therefore, after simplification, we get:

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