The two workhorse uses of induction in the Class-11 syllabus are proving summation formulas and proving divisibility statements. Both follow the same skeleton — base case, hypothesis, step — but the algebra at the inductive step has a characteristic shape worth learning explicitly.
Summation formulas. Here $P(n)$ asserts that a sum of $n$ terms equals a closed formula. The inductive step always starts the same way: write the $(k+1)$-term sum as "the $k$-term sum $+$ the new term", then replace the $k$-term sum using the hypothesis.
$$\underbrace{a_1 + a_2 + \cdots + a_k}_{\text{use hypothesis}} + a_{k+1} \;=\; \big(\text{formula at }k\big) + a_{k+1} \;\overset{?}{=}\; \big(\text{formula at }k+1\big)$$
The whole game is the algebra after the substitution: factor out the common piece (very often $(k+1)$) and rearrange to match the formula with $n$ replaced by $k+1$. The standard results you will prove and reuse are:
A neat consequence of the last two rows: the sum of the first $n$ cubes is the square of the sum of the first $n$ naturals — $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$.
Divisibility statements. Here $P(n)$ claims some expression $f(n)$ is divisible by a fixed integer $d$, written $d \mid f(n)$. The reliable technique is to express $f(k+1)$ in terms of $f(k)$:
$$f(k+1) = f(k) + \big(\text{a multiple of } d\big).$$
By the hypothesis $d \mid f(k)$, and the added piece is a multiple of $d$ by construction, so their sum $f(k+1)$ is also divisible by $d$. The skill is the algebraic manipulation that exposes the "$f(k) + d\cdot(\text{something})$" form — typically by writing the larger power as the previous power times a base, e.g. $7^{k+1} = 7 \cdot 7^k$, then adding and subtracting to recreate $f(k)$. Three classic results:
- $n^3 - n$ is divisible by $6$ for all $n \in \mathbb{N}$.
- $3^{2n} - 1$ is divisible by $8$ for all $n \in \mathbb{N}$.
- $7^n - 3^n$ is divisible by $4$ for all $n \in \mathbb{N}$.
A small but vital habit: keep the divisor visible. Writing $f(k) = d\,m$ for some integer $m$ (rather than just "$f(k)$ is divisible by $d$") makes the final "$= d \times (\text{integer})$" conclusion obvious.
Deeper Insight — both proofs hinge on one identity: $\textit{case}(k+1) = \textit{case}(k) + \textit{new piece}$. Whether you are adding the next term of a sum or peeling one factor off a power, the inductive step succeeds precisely when you can write the $(k+1)$-object as the $k$-object plus something you already control. For sums the "new piece" is the next term; for divisibility it is a deliberate multiple of $d$. This is why a fluent reflex — "how do I see the $k$-case sitting inside the $(k+1)$-case?" — matters more than memorising any single manipulation. Master that reflex and divisibility, summation, and later inequalities all feel like the same proof wearing different clothes.
Prove by induction that $1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: LHS $=1$, RHS $=\dfrac{1\cdot2\cdot3}{6}=1$. True.
- Hypothesis: $1^2+\cdots+k^2=\dfrac{k(k+1)(2k+1)}{6}$.
- Step: add $(k+1)^2$: $\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2=\dfrac{(k+1)\big[k(2k+1)+6(k+1)\big]}{6}$.
- Inside the bracket: $2k^2+7k+6=(k+2)(2k+3)$, so the sum is $\dfrac{(k+1)(k+2)(2k+3)}{6}$.
- This is the formula with $n=k+1$ (since $2(k+1)+1=2k+3$).
Answer: By PMI, $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$ for all $n\in\mathbb{N}$.
Prove by induction that $1^3 + 2^3 + \cdots + n^3 = \left[\dfrac{n(n+1)}{2}\right]^2$.
Solution- Base: $n=1$: LHS $=1$, RHS $=\left[\dfrac{1\cdot2}{2}\right]^2=1$. True.
- Hypothesis: $1^3+\cdots+k^3=\left[\dfrac{k(k+1)}{2}\right]^2=\dfrac{k^2(k+1)^2}{4}$.
- Step: add $(k+1)^3$: $\dfrac{k^2(k+1)^2}{4}+(k+1)^3=\dfrac{(k+1)^2\big[k^2+4(k+1)\big]}{4}=\dfrac{(k+1)^2(k+2)^2}{4}$.
- That equals $\left[\dfrac{(k+1)(k+2)}{2}\right]^2$, the formula at $n=k+1$.
Answer: By PMI, $1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$ for all $n\in\mathbb{N}$.
Prove by induction that the geometric sum $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}$ for $r \ne 1$.
Solution- Base: $n=1$: LHS $=a$, RHS $=\dfrac{a(r-1)}{r-1}=a$. True.
- Hypothesis: $a+ar+\cdots+ar^{k-1}=\dfrac{a(r^k-1)}{r-1}$.
- Step: add $ar^{k}$: $\dfrac{a(r^k-1)}{r-1}+ar^{k}=\dfrac{a(r^k-1)+ar^{k}(r-1)}{r-1}$.
- Numerator $=a\big(r^k-1+r^{k+1}-r^{k}\big)=a\big(r^{k+1}-1\big)$.
- So the sum is $\dfrac{a(r^{k+1}-1)}{r-1}$, the formula at $n=k+1$.
Answer: By PMI, $a+ar+\cdots+ar^{n-1}=\dfrac{a(r^n-1)}{r-1}$ for all $n\in\mathbb{N}$ ($r\ne1$).
Prove by induction that $n^3 - n$ is divisible by $6$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $1-1=0$, divisible by $6$. True.
- Hypothesis: $k^3-k=6m$ for some integer $m$.
- Step: $(k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1=(k^3-k)+3k^2+3k$.
- $=6m+3k(k+1)$. Now $k(k+1)$ is a product of two consecutive integers, hence even, so $3k(k+1)$ is a multiple of $6$.
- Thus $(k+1)^3-(k+1)=6m+6t=6(m+t)$.
Answer: By PMI, $6 \mid (n^3-n)$ for all $n\in\mathbb{N}$.
Prove by induction that $3^{2n} - 1$ is divisible by $8$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $3^2-1=8$, divisible by $8$. True.
- Hypothesis: $3^{2k}-1=8m$ for some integer $m$, i.e. $3^{2k}=8m+1$.
- Step: $3^{2(k+1)}-1=3^{2k}\cdot 9-1=9(8m+1)-1=72m+8=8(9m+1)$.
- A multiple of $8$.
Answer: By PMI, $8 \mid (3^{2n}-1)$ for all $n\in\mathbb{N}$.
Prove by induction that $7^n - 3^n$ is divisible by $4$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $7-3=4$, divisible by $4$. True.
- Hypothesis: $7^k-3^k=4m$ for some integer $m$.
- Step: $7^{k+1}-3^{k+1}=7\cdot7^k-3\cdot3^k=7\cdot7^k-7\cdot3^k+7\cdot3^k-3\cdot3^k$.
- $=7\big(7^k-3^k\big)+(7-3)3^k=7(4m)+4\cdot3^k=4\big(7m+3^k\big)$.
Answer: By PMI, $4 \mid (7^n-3^n)$ for all $n\in\mathbb{N}$.
Prove by induction that $2 + 4 + 6 + \cdots + 2n = n(n+1)$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: LHS $=2$, RHS $=1\cdot2=2$. True.
- Hypothesis: $2+4+\cdots+2k=k(k+1)$.
- Step: add $2(k+1)$: $k(k+1)+2(k+1)=(k+1)(k+2)$.
- This is the formula at $n=k+1$.
Answer: By PMI, $2+4+\cdots+2n=n(n+1)$ for all $n\in\mathbb{N}$.
Prove by induction that $\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\cdots+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}$.
Solution- Base: $n=1$: LHS $=\dfrac{1}{1\cdot2}=\dfrac12$, RHS $=\dfrac{1}{2}$. True.
- Hypothesis: the sum to $k$ terms is $\dfrac{k}{k+1}$.
- Step: add $\dfrac{1}{(k+1)(k+2)}$: $\dfrac{k}{k+1}+\dfrac{1}{(k+1)(k+2)}=\dfrac{k(k+2)+1}{(k+1)(k+2)}$.
- Numerator $=k^2+2k+1=(k+1)^2$, so the sum is $\dfrac{(k+1)^2}{(k+1)(k+2)}=\dfrac{k+1}{k+2}$.
Answer: By PMI, the sum equals $\dfrac{n}{n+1}$ for all $n\in\mathbb{N}$.
Prove by induction that $n(n+1)(n+2)$ is divisible by $6$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $1\cdot2\cdot3=6$, divisible by $6$. True.
- Hypothesis: $k(k+1)(k+2)=6m$ for some integer $m$.
- Step: $(k+1)(k+2)(k+3)=(k+1)(k+2)k+(k+1)(k+2)\cdot3=k(k+1)(k+2)+3(k+1)(k+2)$.
- $=6m+3(k+1)(k+2)$. Since $(k+1)(k+2)$ are consecutive integers, their product is even, so $3(k+1)(k+2)$ is a multiple of $6$.
- Hence the total is a multiple of $6$.
Answer: By PMI, $6 \mid n(n+1)(n+2)$ for all $n\in\mathbb{N}$.
Prove by induction that $1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: LHS $=1$, RHS $=2^1-1=1$. True.
- Hypothesis: $1+2+\cdots+2^{k-1}=2^k-1$.
- Step: add $2^{k}$: $(2^k-1)+2^{k}=2\cdot2^k-1=2^{k+1}-1$.
- This is the formula at $n=k+1$.
Answer: By PMI, $1+2+\cdots+2^{n-1}=2^n-1$ for all $n\in\mathbb{N}$.
Prove by induction that $n^2 + n$ is divisible by $2$ for all $n \in \mathbb{N}$.
Solution- Base: $n=1$: $1+1=2$, divisible by $2$. True.
- Hypothesis: $k^2+k=2m$ for some integer $m$.
- Step: $(k+1)^2+(k+1)=k^2+2k+1+k+1=(k^2+k)+2(k+1)=2m+2(k+1)=2(m+k+1)$.
Answer: By PMI, $2 \mid (n^2+n)$ for all $n\in\mathbb{N}$.