Apply part (b) repeatedly. After the first puncturing, the code has rank $k-1$, length $n-d$, and minimum distance at least $\lceil d/2\rceil$. Repeating with a minimum-weight codeword at each stage gives
$$
n\geq d+\left\lceil\frac d2\right\rceil
+\left\lceil\frac d{2^2}\right\rceil+cdots
+\left\lceil\frac d{2^{k-1}}\right\rceil,
$$
because nested ceilings obey
$$
\left\lceil\frac{\lceil d/2^j\rceil}{2}\right\rceil
=\left\lceil\frac d{2^{j+1}}\right\rceil.
$$
Thus
$$
\boxed{n\geq d+\sum_{1\leq l\leq k-1}
\left\lceil\frac d{2^l}\right\rceil},
$$
which is the binary <Griesmer bound>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 11G / c / Solution

{c}
{wiki}

Every binary linear $[n,k,d]$ code satisfies
$$
n\geq\sum_{j=0}^{k-1}\left\lceil\frac d{2^j}\right\rceil.
$$
Puncturing on the support of a minimum-weight codeword reduces the rank by one and leaves minimum distance at least $\lceil d/2\rceil$, which proves the bound inductively.


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