Here is a different, perhaps more elementary example.

Consider the Grassmannian $\mathrm{Gr}(2,4)$ of lines in $\mathbf{P}^3$, and let $\mathscr{Q}$ be the universal quotient bundle. The line bundle $\mathrm{det}(\mathscr{Q})$ generates the Picard group of $\mathrm{Gr}(2,4)$. If the rank $2$ bundle $\mathscr{Q}$ were in the subgroup of $\mathrm{K}^0$ generated by line bundles, then its K-theory class would therefore have to be $2[\mathrm{det}(\mathscr{Q})]$. Hence $c(\mathscr{Q})=c(\mathrm{det}(\mathscr{Q}))^2$ and 
$$c_2(\mathscr{Q})=c_1(\mathscr{Q})^2,$$
which cannot be true. Indeed, fix a point $p$ and a line $L$ in $\mathbf{P}^3$; the Chern class $c_1(\mathscr{Q})$ is geometrically represented by the Schubert cycle $\Sigma_1(L)=\{L' \ | \ L\cap L'\neq\emptyset\}$, while $c_2(\mathscr{Q})$ is represented by $\Sigma_2(p)=\{L' \ | \ p\in L'\}$. Then
$$1=\int_{\mathrm{Gr}(2,4)} c_2(\mathscr{Q})^2\neq\int_{\mathrm{Gr}(2,4)} c_1(\mathscr{Q})^4=2.$$
The integral on the left is the number of lines through two distinct points in $\mathbf{P}^3$, while the integral on the right is the number of lines meeting four given (general) lines in $\mathbf{P}^3$.