The inequality \[(n+1) + \bigl(n^4+2\bigr) + \bigl(n^9+3\bigr) +
\bigl(n^{16}+4\bigr) + \cdots + \bigl(n^{10000}+100\bigr) > k\] is true for all \(n\ge 1\). It follows that
\(k < 1300\),
\(k^2<101\),
\(k \ge 101^{10000}\),
\(k < 5150\).
To make this a lot simpler, note that the expression
\[(n+1) + \bigl(n^4+2\bigr) + \bigl(n^9+3\bigr) +
\bigl(n^{16}+4\bigr) + \cdots + \bigl(n^{10000}+100\bigr)\] increases as
\(n\) increases, so the inequality holds for all
\(n \ge 1\) if it holds for
\(n = 1\). So we need
\[\begin{align*}
&(1+1)+(1+2)+(1+3)+(1+4)+\cdots +(1+100) > k\\
\iff\quad& 2+3+4+5+\cdots +101 > k\\
\iff\quad& \frac{100}{2}(2+101) > k\\
\iff\quad& 5150 > k.
\end{align*}\]
Hence the answer is (d).
Alternatively, we could have broken up the sum here as \[(1+1+\cdots+1)+(1+2+\cdots+100)=100+\tfrac{1}{2}\times100\times101.\]
