CS3000: Problem Set 0 — Typesetting Practice & Math Refresher
Complete Solutions
Shahid Shaikh
September 2025
Problem 1: Proof Practice
1.
n
X
i=1
i2 =
n(n + 1)(2n + 1)
6
Proof (by induction).
P1 2
1·2·3
Base case n = 1.
i=1 i = 1. The right-hand side (RHS) is 6 = 1.True for n = 1.
Inductive step. Assume that for some k ∈ N,
k+1
X
i2 =
i=1
k
X
Pk
i=1 i
2 = k(k+1)(2k+1) . Then
6
i2 + (k + 1)2
i=1
k(k + 1)(2k + 1)
+ (k + 1)2
6
(k + 1) k(2k + 1) + 6(k + 1)
=
6
(k + 1)(2k 2 + 7k + 6)
=
6
(k + 1)(k + 2)(2k + 3)
.
=
6
(k+1) (k+1)+1 2(k+1)+1
This equals
, which is the RHS with n = k + 1. Therefore, by induction,
6
the identity holds for all n ∈ N.
=
Problem 2: Passage (Typesetting Text)
Typeset your favorite passage from a book, movie, series, song, game; anywhere basically. Just let
me know where it is from as well.
“Discipline equals freedom. Without discipline, you don’t have freedom.”
— Joe Rogan, JRE Podcast
1
Problem 3: Miscellaneous Mathematics
a) Fibonacci numbers. Define F0 = 0, F1 = 1, and for n ≥ 2, Fn = Fn−1 + Fn−2 .
√
−b ± b2 − 4ac
2
b) Roots of ax + bx + c = 0. For a ̸= 0, the solutions are x =
.
2a
Problem 4: Including Images
Solution:
Figure 1: T-Rex.
2