Sum of First n Natural Numbers

$\sum_{k=1}^{n} k = \frac{1}{2}n(n+1)$

Sum of Squares

$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Sum of Cubes

$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$

Example: Find the sum of squares of the first 10 natural numbers.

Solution: $\sum_{k=1}^{10} k^2 = \frac{10(10+1)(2\times10+1)}{6} = \frac{10 \times 11 \times 21}{6} = 385$

Solving Quadratic Inequalities

1. Write the inequality in standard form (zero on one side)
2. Factor the quadratic expression
3. Use a sign chart to determine where the expression is positive or negative

Example: Solve $x^2 + x - 6 > 0$

Solution: $(x - 2)(x + 3) > 0$

Using the sign chart:

Therefore, the solution is $x < -3$ or $x > 2$.

Solving Rational Inequalities

1. Make one fraction on the left side
2. Find critical values (where numerator or denominator equals zero)
3. Test intervals with a sign chart

Example: Solve $\frac{2x+1}{x} > 1$

Solution: $\frac{2x+1}{x} - 1 > 0 \Rightarrow \frac{x+1}{x} > 0$

Critical values: $x = -1$ (numerator) and $x = 0$ (denominator)

Solution: $x < -1$ or $x > 0$

Determinant of a 3×3 Matrix

For $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$:

$|A| = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

Inverse of a 3×3 Matrix

$A^{-1} = \frac{1}{|A|} \cdot \text{Adj}(A)$

where Adj(A) is the transpose of the cofactor matrix.

Example: Find the determinant of $A = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 5 & 3 \end{pmatrix}$

Solution: $|A| = 1\begin{vmatrix}1 & 1 \\ 5 & 3 \end{vmatrix} - 2\begin{vmatrix}1 & 1 \\ 2 & 3 \end{vmatrix} + 1\begin{vmatrix}1 & 1 \\ 2 & 5 \end{vmatrix}$ $= 1(3-5) - 2(3-2) + 1(5-2)$ $= -2 - 2 + 3 = -1$

Decomposition Method

Example: Decompose $\frac{6x}{(x-1)(x+2)}$

Solution: $\frac{6x}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

$6x = A(x+2) + B(x-1)$

Setting $x = 1$: $6 = 3A \Rightarrow A = 2$ Setting $x = -2$: $-12 = -3B \Rightarrow B = 4$

Therefore: $\frac{6x}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{4}{x+2}$

The principle of mathematical induction proves statements for all natural numbers $n$:

1. Base Case: Prove true for $n = 1$
2. Inductive Hypothesis: Assume true for $n = k$
3. Inductive Step: Prove true for $n = k + 1$

Example: Prove by induction that $\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$

Proof:

Step 1: For $n = 1$, LHS = 1, RHS = $\frac{1}{2}(1)(2) = 1$ ✓

Step 2: Assume true for $n = k$: $1 + 2 + ... + k = \frac{1}{2}k(k+1)$

Step 3: Prove for $n = k + 1$: $1 + 2 + ... + k + (k+1) = \frac{1}{2}k(k+1) + (k+1)$ $= \frac{k(k+1) + 2(k+1)}{2} = \frac{1}{2}(k+1)(k+2)$

Thus true for $n = k + 1$. Therefore, true for all $n \in \mathbb{N}$.

For any positive integer $n$: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

General Term

$T_{r+1} = \binom{n}{r} a^{n-r} b^r$

Example: Find the middle term in $(2x + 3)^6$

Solution: Here $n = 6$ (even), so the middle term is the $\frac{6}{2} + 1 = 4^{th}$ term.

$T_4 = \binom{6}{3} (2x)^3 (3)^3 = 20 \times 8x^3 \times 27 = 4320x^3$

Binomial Expansion for Fractional/Negative Indices

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$

This expansion is valid for $|x| < 1$.

Example: Expand $(1 + 2x)^{-1}$ up to $x^3$

Solution: $(1 + 2x)^{-1} = 1 + (-1)(2x) + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3 + ...$ $= 1 - 2x + 4x^2 - 8x^3 + ...$

Valid for $|2x| < 1$, i.e., $|x| < \frac{1}{2}$

In Tanzania, these algebraic concepts are used in various practical situations. For example, when a shopkeeper in Dar es Salaam wants to determine the optimal pricing for their goods to maximize profit, they can use quadratic inequalities to find the range of prices that yield desired profit margins. A small business owner can use matrices to solve systems of linear equations when calculating costs involving multiple products and expenses—for instance, determining how many units of different agricultural products (maize, beans, rice) to stock given constraints on storage space and budget, using matrix inverse methods to solve the resulting system efficiently.