Mathematical Induction

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Principle of Mathematical Induction (PMI): To prove a statement $P(n)$ for all natural numbers $n \geq 1$:

1. Base Case: Prove $P(1)$ is true.
2. Inductive Hypothesis: Assume $P(k)$ is true for some arbitrary $k \geq 1$.
3. Inductive Step: Using the hypothesis, prove $P(k+1)$ is true.

If all three steps succeed, then $P(n)$ is true for all $n \geq 1$.

Standard Sum Formulas (proveable by PMI): $$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$ $$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2$$

Common Applications: Proving divisibility (e.g., $6^n - 1$ is divisible by 5), inequality bounds (e.g., $2^n > n^2$ for $n \geq 5$), and summation identities.

⚡ Exam tip: For JEE Advanced, the inductive step often involves algebraic manipulation — factorising $P(k+1) - P(k)$ or rewriting terms cleverly.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and problem-solving strategies.

Why Induction Works

Mathematical induction is valid because of the well-ordering principle: every non-empty set of natural numbers has a least element. If $P(n)$ is false for some $n$, let $m$ be the smallest such $n$. Then $P(m)$ is false but $P(m-1)$ is true, violating the inductive step. Thus no counterexample exists.

Strong Induction Variant:

Instead of assuming $P(k)$, we assume $P(1), P(2), …, P(k)$ are all true (strong hypothesis), then prove $P(k+1)$. Strong induction is equivalent to ordinary induction but is more convenient when $P(k+1)$ depends on several earlier cases.

Example — Sum of First n Numbers:

Base Case ($n=1$): LHS $= 1$, RHS $= \frac{1(2)}{2} = 1$. ✓

Inductive Step: Assume $\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$.

Then $\sum_{r=1}^{k+1} r = \left(\sum_{r=1}^{k} r\right) + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$.

But $\frac{(k+1)(k+2)}{2}$ is exactly $\frac{(k+1)((k+1)+1)}{2}$, which is $P(k+1)$. ✓

Example — Sum of Squares:

Base Case ($n=1$): $1^2 = \frac{1(2)(3)}{6} = 1$. ✓

Inductive Step: Assume $\sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6}$.

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

$$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6} = \frac{(k+1)[2k^2 + k + 6k + 6]}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6}$$

$$2k^2 + 7k + 6 = (k+2)(2k+3) = (2k+3)(k+2)$$

So $= \frac{(k+1)(k+2)(2k+3)}{6} = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$. ✓

Divisibility Proofs:

Example: Prove $6^n - 1$ is divisible by 5 for all $n \geq 1$.

Base Case ($n=1$): $6^1 - 1 = 5$, divisible by 5. ✓

Inductive Step: Assume $5 | (6^k - 1)$, i.e., $6^k - 1 = 5m$ for some integer $m$.

$$6^{k+1} - 1 = 6 \cdot 6^k - 1 = 6(6^k - 1) + 6 - 1 = 6(5m) + 5 = 5(6m + 1)$$

Since $6m + 1$ is an integer, $5 | (6^{k+1} - 1)$. ✓

Example: Prove $11^n - 6$ is divisible by 5 for all $n \geq 1$.

Base Case ($n=1$): $11 - 6 = 5$, divisible by 5. ✓

Inductive Step: Assume $5 | (11^k - 6)$, so $11^k - 6 = 5m$.

$$11^{k+1} - 6 = 11 \cdot 11^k - 6 = 11(11^k - 6) + 66 - 6 = 11(5m) + 60 = 5(11m + 12)$$

Since $11m + 12$ is an integer, $5 | (11^{k+1} - 6)$. ✓

Inequality Proofs:

Example: Prove $2^n > n^2$ for all $n \geq 5$.

Base Case ($n=5$): $2^5 = 32$, $5^2 = 25$. $32 > 25$. ✓

Inductive Step: Assume $2^k > k^2$ for some $k \geq 5$.

$$2^{k+1} = 2 \cdot 2^k > 2 \cdot k^2$$

We need $2k^2 > k^2 + 2k + 1$, i.e., $k^2 - 2k - 1 > 0$, i.e., $(k-1)^2 > 2$.

For $k \geq 3$: $(k-1)^2 \geq 4 > 2$. So $2k^2 > k^2 + 2k + 1$ holds for $k \geq 3$.

Since $k \geq 5 > 3$, we have $2^{k+1} = 2 \cdot 2^k > 2k^2 > k^2 + 2k + 1 = (k+1)^2$. ✓

JEE Advanced Worked Example:

Prove by induction: $\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}$.

Base Case ($n=1$): LHS $= 1 \cdot 2 = 2$, RHS $= \frac{1 \cdot 2 \cdot 3}{3} = 2$. ✓

Inductive Step: Assume $\sum_{r=1}^{k} r(r+1) = \frac{k(k+1)(k+2)}{3}$.

$$\sum_{r=1}^{k+1} r(r+1) = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3}$$

$$= \frac{(k+1)(k+2)(k+3)}{3} = \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$$

This is $P(k+1)$. ✓

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🔴 Extended — Deep Study (3mo+)

Comprehensive derivations, advanced applications, previous year patterns, and sophisticated problem types.

Strong Induction — Fibonacci Identity

The Fibonacci sequence is defined by $F_1 = 1, F_2 = 1, F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$.

Identity: $F_1^2 + F_2^2 + \cdots + F_n^2 = F_n \cdot F_{n+1}$.

Base Cases:

• $n=1$: LHS $= 1^2 = 1$, RHS $= F_1 \cdot F_2 = 1 \cdot 1 = 1$. ✓
• $n=2$: LHS $= 1^2 + 1^2 = 2$, RHS $= F_2 \cdot F_3 = 1 \cdot 2 = 2$. ✓

Inductive Step: Assume $F_1^2 + \cdots + F_k^2 = F_k F_{k+1}$ for all $k \leq n$.

For $n+1$: $$F_1^2 + \cdots + F_n^2 + F_{n+1}^2 = F_n F_{n+1} + F_{n+1}^2 = F_{n+1}(F_n + F_{n+1}) = F_{n+1} F_{n+2}$$

Since $F_{n+2} = F_{n+1} + F_n$ (definition). ✓

Matrix Identity by Induction:

Claim: For any $n \geq 1$, if $A = \begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix}$, then $A^n = \begin{pmatrix} F_{n+1} & F_n \ F_n & F_{n-1} \end{pmatrix}$.

Base Case ($n=1$): $A^1 = \begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_2 & F_1 \ F_1 & F_0 \end{pmatrix}$. Taking $F_0 = 0$, this holds. ✓

Inductive Step: Assume $A^k = \begin{pmatrix} F_{k+1} & F_k \ F_k & F_{k-1} \end{pmatrix}$.

$$A^{k+1} = A^k \cdot A = \begin{pmatrix} F_{k+1} & F_k \ F_k & F_{k-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_{k+1}+F_k & F_{k+1} \ F_k+F_{k-1} & F_k \end{pmatrix}$$

Since $F_{k+2} = F_{k+1} + F_k$ and $F_{k+1} = F_k + F_{k-1}$: $$= \begin{pmatrix} F_{k+2} & F_{k+1} \ F_{k+1} & F_k \end{pmatrix} = \begin{pmatrix} F_{(k+1)+1} & F_{k+1} \ F_{k+1} & F_{(k+1)-1} \end{pmatrix}$$

This is $P(k+1)$. ✓

Determinant Identity by Induction:

Claim: $\begin{vmatrix} 2\cos\theta & 1 & 0 \ 1 & 2\cos\theta & 1 \ 0 & 1 & 2\cos\theta \end{vmatrix} = \frac{\sin 4\theta}{\sin\theta}$.

This type of tridiagonal determinant satisfies a recurrence relation. The determinant $D_n$ for an $n \times n$ tridiagonal matrix with $2\cos\theta$ on the diagonal and $1$ on the off-diagonals satisfies $D_n = 2\cos\theta \cdot D_{n-1} - D_{n-2}$ with $D_0 = 1, D_1 = 2\cos\theta$.

By induction, $D_n = \frac{\sin((n+1)\theta)}{\sin\theta}$. For $n=3$: $D_3 = \frac{\sin 4\theta}{\sin\theta}$. ✓

Cauchy Induction (Rearrangement Inequality):

The rearrangement inequality states that for two sequences sorted in the same order, the sum of products is maximum. We can use induction on the number of terms.

Divisibility by 3 and 11 — Combined Pattern:

$n^3 - n$ is divisible by 6 (and hence by 3): $n^3 - n = n(n-1)(n+1)$, product of three consecutive integers, always divisible by 3. Also divisible by 2 (at least one even factor). So divisible by 6.

Prove $11^n + 2$ is divisible by 3:

• Base: $n=0$: $11^0 + 2 = 3$, divisible by 3. ✓
• Step: Assume $3 | (11^k + 2)$. Then $11^{k+1} + 2 = 11 \cdot 11^k + 2 = 11(11^k + 2) - 22 + 2 = 11(11^k + 2) - 20$. Since $3 | (11^k + 2)$ and $3 | 20$? Actually, $-20$ is not divisible by 3. Let me redo: $11^{k+1} + 2 = 11(11^k + 2) - 22 + 2 = 11(11^k + 2) - 20 = 11(11^k + 2) - 18 - 2 = 11(11^k + 2) - 3(6) - 2$. Hmm…

Better approach: $11 \equiv 2 \pmod{3}$, so $11^n \equiv 2^n \pmod{3}$. $2^n \equiv 2$ when $n$ is odd, $2^n \equiv 1$ when $n$ is even. Actually: $11^n + 2 \equiv 2^n + 2 \pmod{3}$. For $n=1$: $2+2=4\equiv 1$. For $n=2$: $4+2=6\equiv 0$. For $n=3$: $8+2=10\equiv 1$. So $11^n+2$ is divisible by 3 iff $n$ is even.

By induction: $11^n + 2 = 11(11^{n-1} + 2) - 20$. If $n-1$ is even, $11^{n-1} + 2 \equiv 1+2 \equiv 0 \pmod{3}$. Then $11(0) - 20 \equiv -20 \equiv -2 \not\equiv 0$. The inductive step is easier directly: $11^{n+2} + 2 = 121 \cdot 11^n + 2 \equiv 1 \cdot 11^n + 2 \equiv 1 \cdot (11^n + 2) \pmod{3}$.

De Moivre’s Theorem by Induction:

$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$.

For natural number $n$, this follows by induction on $n$ using the multiplication formula for complex numbers.

JEE Advanced Previous Year Patterns:

Topic	Frequency (2015–2024)
Sum formula proofs (Σn, Σn², Σn³)	1 question per 2 years
Divisibility proofs (5, 7, 11, etc.)	1 question per 2 years
Inequality proofs	1 question per 3 years
Matrix/determinant identities	1 question per 4 years
Fibonacci identities	1 question per 3 years
Sum of odd numbers / square numbers	1 question per 3 years

Advanced Worked Example (JEE 2017): Prove that $\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^n} = 1 - \frac{1}{2^n}$ for all $n \geq 1$.

Base Case ($n=1$): LHS $= \frac{1}{2}$, RHS $= 1 - \frac{1}{2} = \frac{1}{2}$. ✓

Inductive Step: Assume $\sum_{r=1}^{k} \frac{1}{2^r} = 1 - \frac{1}{2^k}$.

$$\sum_{r=1}^{k+1} \frac{1}{2^r} = \left(1 - \frac{1}{2^k}\right) + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{2}{2^{k+1}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}$$

This is $P(k+1)$. ✓

Geometric Progression Sum by Induction:

Claim: $1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1$.

Base Case ($n=1$): $1 = 2^1 - 1 = 1$. ✓

Inductive Step: Assume $1 + 2 + \cdots + 2^{k-1} = 2^k - 1$.

$1 + 2 + \cdots + 2^{k-1} + 2^k = (2^k - 1) + 2^k = 2 \cdot 2^k - 1 = 2^{k+1} - 1$. ✓

Multinomial Coefficient Sum:

Prove $\sum_{r=0}^{n} \binom{n}{r} = 2^n$.

Base Case ($n=0$): $\binom{0}{0} = 1 = 2^0$. ✓

Inductive Step: Using Pascal’s identity $\binom{k+1}{r} = \binom{k}{r} + \binom{k}{r-1}$: $$\sum_{r=0}^{k+1} \binom{k+1}{r} = \sum_{r=0}^{k+1}\binom{k}{r} + \sum_{r=0}^{k+1}\binom{k}{r-1} = \sum_{r=0}^{k}\binom{k}{r} + \sum_{r=1}^{k+1}\binom{k}{r-1}$$ $$= 2^k + \sum_{s=0}^{k}\binom{k}{s} = 2^k + 2^k = 2^{k+1}$$

Alternatively, using the binomial theorem: $(1+1)^n = \sum_{r=0}^{n}\binom{n}{r}$. But induction gives the same result.

Cauchy-Schwarz Inequality (Finite Form):

$$\left(\sum_{i=1}^{n} a_i b_i\right)^2 \leq \left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)$$

Can be proved by induction on $n$ using the identity for $n=2$ and showing adding the $(n+1)$th term preserves the inequality.

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