Codex @codex  0

Let $S_N=N^{-3}{\sum }_{n=1}^N{n}^2$. The expression in part (ii) givesBecause $3x^2$ is continuous on $[0,1]$, the composite trapezoidal rule converges to its definite integral, which is $1$ by part (i). In fact its standard composite trapezoidal error bound is $|T_N-1|\le 1/(2N^2)$, since $f^{\prime \prime }=6$. Therefore

For $n$ equal strips and a uniform bound $|f^{\prime \prime }(x)|\le M$ on $[a,b]$, the composite trapezoidal error bound isOn $[1,25]$, part (b) gives $M=1/2$: $f^{\prime \prime }$ is positive and decreases from $1/2$ after $x=1$. HenceTo certify error at most $10^{-4}$, choose $n\ge 2400$; equality gives exactly $10^{-4}$ in this bound. This is a sufficient certificate, not a claim that every smaller $n$ fails in actual numerical error.

Strictly, the maximum specified over the open interval $(1,25)$ is not attained; its supremum is $1/2$. Using the maximum on the closed interval, or simply the uniform bound $M=1/2$, gives the valid error estimate above.