A statistician needs a closed-form expression for $1 + 2 + 3 + \cdots + n$, the formula $\dfrac{n(n+1)}{2}$ is "obvious" from Gauss's playground trick, but how do we prove it holds for every positive integer? Mathematical induction is the answer, and the technique is mechanical once you internalise the algebra: add the $(k+1)$-th term, then factor the right-hand side until $\dfrac{(k+1)(k+2)}{2}$ pops out.
Start with the warm-up and record your first reasoning.
You want to prove $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$. Before any factoring when you assume the formula at $n = k$, what is the very next algebraic move? And what does the right-hand side need to look like at the end for the proof to close?
Orient yourself to the lesson goals, language, and core approach.
Every series induction proof rewards two habits: add the next term to both sides of the hypothesis, then factor the RHS to match the target form at $n = k+1$. The target form is exactly what you get by replacing $k$ with $k+1$ in the closed expression, write it down before you start factoring so you know where you are heading.
The add-and-factor strategy: (1) state $S_k = \dots$ as the hypothesis, (2) compute $S_{k+1} = S_k + a_{k+1}$ (add the next term), (3) factor the RHS until it matches the formula evaluated at $k+1$.
Target: $S_{k+1} = \dfrac{(k+1)(k+2)}{2}$ · Found: $\dfrac{k(k+1)}{2} + (k+1) = \dfrac{(k+1)(k+2)}{2}$ ✓
Work through the concept and complete its check.
Every series induction problem follows the same four-step skeleton:
Worked through the hook, prove $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$:
Four steps: base, hypothesis, step ($+a_{k+1}$ and factor), conclude · Always write the target form (formula at $n=k+1$) before factoring · $\sum i = \frac{n(n+1)}{2}$: factor $(k+1)$ then simplify $\left[\frac{k}{2}+1\right] = \frac{k+2}{2}$ · Closing line: "by induction, $P(n)$ true for all integers $n \geq 1$"
Pause, copy the series induction four steps (base, hypothesis, $+a_{k+1}$ and factor, conclude) and the closing phrase "by induction, $P(n)$ true for all $n \geq 1$" into your book.
Quick check: In the inductive step for $\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$, after adding $(k+1)^2$ to both sides, what is the target form you need to factor towards?
Build the next proof move and complete its check.
We just saw that the series induction step adds $a_{k+1}$ to both sides and factors $(k+1)$ from the numerator to match the formula at $n = k+1$. That raises a question: is there a non-inductive way to sum certain series, and when is it cleaner? This card answers it → telescoping sums $\sum(b_i - b_{i-1}) = b_n - b_0$, where partial fractions first convert the summand.
If you can write the $i$-th term as a difference $a_i = b_i - b_{i-1}$, the sum collapses: $\sum_{i=1}^{n}(b_i - b_{i-1}) = b_n - b_0$. This is called telescoping. It's both a stand-alone proof technique and a useful way to sanity-check an induction target.
Example: Prove $\displaystyle\sum_{i=1}^{n} \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}$.
You can also prove this by induction, the step is $S_{k+1} = \dfrac{k}{k+1} + \dfrac{1}{(k+1)(k+2)} = \dfrac{k(k+2) + 1}{(k+1)(k+2)} = \dfrac{k^2 + 2k + 1}{(k+1)(k+2)} = \dfrac{(k+1)^2}{(k+1)(k+2)} = \dfrac{k+1}{k+2}$. ✓
Telescoping: $\sum_{i=1}^{n}(b_i - b_{i-1}) = b_n - b_0$ · $\dfrac{1}{i(i+1)} = \dfrac{1}{i} - \dfrac{1}{i+1}$, partial fractions before telescoping · $\sum \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}$, provable by telescoping OR induction · Always write out first 3 and last 2 terms to confirm cancellation pattern
Pause, copy the telescoping identity $\sum_{i=1}^n (b_i - b_{i-1}) = b_n - b_0$, the partial-fraction split $\frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1}$, and the result $\sum \frac{1}{i(i+1)} = \frac{n}{n+1}$ into your book.
Did you get this? True or false: the inductive step for $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$ requires the algebraic simplification $\frac{k(k+2)+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$.
Follow the worked examples, then complete the check.
Worked examples · 3 in a row, reveal as you go
Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$ for all positive integers $n$.
Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$.
Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} i^3 = \left(\dfrac{n(n+1)}{2}\right)^2 = \dfrac{n^2(n+1)^2}{4}$ for all positive integers $n$.
Fill the gap: When proving $\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}$ by induction, after factoring $(k+1)^2$ from $\frac{k^2(k+1)^2}{4} + (k+1)^3$, the remaining bracket simplifies to $k^2 + 4(k+1) = (k+m)^2$, where $m =$ .
Correct the common traps and test your understanding.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when proving $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$, factoring $(k+1)$ first gives $\frac{k(k+1)}{2} + (k+1) = (k+1)\left[\frac{k}{2} + 1\right] = (k+1)\cdot \frac{k+2}{2} = \frac{(k+1)(k+2)}{2}$.
Complete the practice activities and revisit your opening thinking.
Activities · practice with the ideas
Prove by induction that $\displaystyle\sum_{i=1}^{n}(2i - 1) = n^2$ (sum of first $n$ odd numbers).
Prove by induction that $\displaystyle\sum_{i=1}^{n} i(i+1) = \dfrac{n(n+1)(n+2)}{3}$.
Prove by induction that $\displaystyle\sum_{i=1}^{n} \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}$. (Alternatively, prove by telescoping.)
Prove by induction that $\displaystyle\sum_{i=1}^{n} (2i - 1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$ (sum of squares of first $n$ odd numbers).
Use telescoping to evaluate $\displaystyle\sum_{i=1}^{n} \left[\dfrac{1}{i} - \dfrac{1}{i+1}\right]$, then state the result in closed form and verify it agrees with the formula from drill 3.
Odd one out: Three of these statements about the inductive proof of $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ are correct. Which one is NOT?
Earlier you sketched the inductive step for $\sum i = \dfrac{n(n+1)}{2}$, what you add, and what target form you need to reach.
You add $(k+1)$ to both sides (it's the next term of the series). The target form is $\dfrac{(k+1)(k+2)}{2}$, exactly what you get by replacing $n$ with $k+1$ in the formula. The factoring trick is to pull $(k+1)$ out: $\dfrac{k(k+1)}{2} + (k+1) = (k+1) \cdot \dfrac{k+2}{2}$. Writing the target form down before factoring is the single biggest discipline improvement for this style of proof.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$ for all positive integers $n$. (2 marks)
Q2. Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. (3 marks)
Q3. Prove by mathematical induction that $\displaystyle\sum_{i=1}^{n} \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}$ for all positive integers $n$. (3 marks)
Activity answers:
1. Base $n=1$: $1=1^2$ ✓. Assume $\sum_{i=1}^{k}(2i-1)=k^2$. $S_{k+1}=k^2+(2(k+1)-1)=k^2+2k+1=(k+1)^2$. ∎
2. Base $n=1$: $1\cdot 2=2$, $\frac{1\cdot 2\cdot 3}{3}=2$ ✓. Assume $S_k=\frac{k(k+1)(k+2)}{3}$. $S_{k+1}=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=(k+1)(k+2)\left[\frac{k}{3}+1\right]=\frac{(k+1)(k+2)(k+3)}{3}$. ∎
3. Base $n=1$: $\frac{1}{1\cdot 2}=\frac{1}{2}$; $\frac{1}{2}$ ✓. Assume $S_k=\frac{k}{k+1}$. $S_{k+1}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}$. ∎
4. Base $n=1$: $1^2=1$; $\frac{1\cdot 1\cdot 3}{3}=1$ ✓. Assume $S_k=\frac{k(2k-1)(2k+1)}{3}$. $S_{k+1}=\frac{k(2k-1)(2k+1)}{3}+(2k+1)^2 = (2k+1)\left[\frac{k(2k-1)}{3}+(2k+1)\right] = \frac{(2k+1)[k(2k-1)+3(2k+1)]}{3} = \frac{(2k+1)(2k^2+5k+3)}{3} = \frac{(2k+1)(k+1)(2k+3)}{3} = \frac{(k+1)(2k+1)(2k+3)}{3}$. ∎
5. Telescoping: terms cancel pairwise, leaving $1-\frac{1}{n+1}=\frac{n}{n+1}$. Agrees with drill 3 closed form. ✓
Q1 (2 marks): Base ✓ [1]. Assume; $S_{k+1}=\frac{k(k+1)}{2}+(k+1)=\frac{(k+1)(k+2)}{2}$. By induction. [1].
Q2 (3 marks): Base ✓ [1]. Assume; $S_{k+1}=\frac{k(k+1)(2k+1)}{6}+(k+1)^2 = \frac{(k+1)[k(2k+1)+6(k+1)]}{6}$ [1] $= \frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$. By induction. [1].
Q3 (3 marks): Base ✓ [1]. Assume $S_k=\frac{k}{k+1}$ [1]. $S_{k+1}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}$. By induction. [1].
Sit the module quiz, then close the lesson off.
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