I2.1 By considering: $\frac{1^{2}}{1}=1, \frac{1^{2}+2^{2}}{1+2}=\frac{5}{3}, \frac{1^{2}+2^{2}+3^{2}}{1+2+3}=\frac{7}{3}, \frac{1^{2}+2^{2}+3^{2}+4^{2}}{1+2+3+4}=3$, find $a$ such that $\frac{1^{2}+2^{2}+\cdots+a^{2}}{1+2+\cdots+a}=\frac{25}{3}$

I2.2 A triangular pyramid is cut from a corner of a cube with side length $a \mathrm{~cm}$ as the figure shown. If the volume of the pyramid is $b \mathrm{~cm}^{3}$, find $b$.

I2.3 If the value of $x^{2}+c x+b$ is not less than 0 for all real number $x$, find the maximum value of $c$
I2.4 If the unit digit of $1997^{1997}$ is $c-d$, find $d$.
The given is equivalent to: $\frac{1^{2}}{1}=\frac{3}{3}, \frac{1^{2}+2^{2}}{1+2}=\frac{5}{3}, \frac{1^{2}+2^{2}+3^{2}}{1+2+3}=\frac{7}{3}, \frac{1^{2}+2^{2}+3^{2}+4^{2}}{1+2+3+4}=\frac{9}{3}$ and $2 \times 1+1=3,2 \times 2+1=5,2 \times 3+1=7,2 \times 4+1=9$; so $2 a+1=25$
$$
\begin{array}{l}
\Rightarrow a=12 \\
b=\frac{1}{3} \text { base area } \times \text { height }=\frac{1}{3}\left(\frac{\frac{1}{2} a \times \frac{1}{2} a}{2}\right) \times \frac{1}{2} a \\
\quad=\frac{1}{48} a^{3}=\frac{1}{48} \cdot 12^{3}=36 \\
x^{2}+c x+36 \geq 0 \\
\Delta=c^{2}-4(36) \leq 0 \\
\Rightarrow c \leq 12
\end{array}
$$

The maximum value of $c=12$.
$$
1997^{1997} \equiv 7^{1997} \equiv 7^{4(499)+1} \equiv 7(\bmod 10)
$$

The unit digit of $1997^{1997}$ is 7
$$
\begin{array}{l}
12-d=7 \\
d=5
\end{array}
$$