= Solution
The leading error term can be removed by comparing the two step sizes $h$ and $h/2$:
$$
I_h=I-K_2h^2-K_4h^4-\cdots,\qquad
I_{h/2}=I-\frac14K_2h^2-\frac1{16}K_4h^4-\cdots.
$$
Multiply the second expression by four and subtract the first. The $h^2$ terms cancel, so the first <Richardson extrapolation> step is
$$
\boxed{R_1(h)=\frac{4I_{h/2}-I_h}{3},\qquad I-R_1(h)=-\frac14K_4h^4+O(h^6).}
$$
Repeat with $R_1(h/2)$ and $R_1(h)$, now cancelling the fourth-order error using the factor $16$. In general, with $R_0(h)=I_h$,
$$
\boxed{R_j(h)=\frac{4^jR_{j-1}(h/2)-R_{j-1}(h)}{4^j-1}.}
$$
Under the stated even-power expansion, successive columns have errors of orders $h^4,h^6,h^8,\ldots$. Each step uses the known scaling of the leading error to eliminate its unknown coefficient.
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