Principle of Mathematical Induction Practice Questions

Mathematical Induction is a proof technique used to show that a mathematical statement is true for all natural numbers (1, 2, 3, ...). It works by first proving the statement for the first natural number (base case) and then proving that if it is true for one number, it is also true for the next number (induction step).

Question 1: For all n ≥ 1, prove that, 1² + 2² + 3²+....+n² = {n(n + 1) (2n + 1)} / 6

Solution:

Let the given statement be P(n),

For n = 1, the left side :
1 ² = 1

The right-hand side:
P(n):1^2+ 2^2 + 3^2+ \ldots+ n^2 = \frac{n(n + 1) (2n + 1)}{6} \\~\\ \text{For n=1} \\ P(1):\frac{1(1+1)(2×1+1)}{6} = 1        

Now, let's take a positive integer, k, and assume P(k) to be true i.e.,

1^2 + 2^2 + 3^2 +....+k^2 = \frac{k(k+1)(2k+1)}{6}        

We shall now prove that P(k + 1) is also true, so now we have,

P(k + 1) = P(k) + (k + 1)²  

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2
= (k+1) \frac{( 2k^2 + k) + 6(k+1)}{6}
=\frac{(k+1)(2k^2 +7k+6)}{6}
=\frac{(k+1) (k+2) (2k+3)}{6}
=\frac{(k+1) ((k+1)+1) (2(k+1) +1)}{6}

Thus P(k + 1) is true, whenever P(k) is true for all natural numbers. Hence, by the process of mathematical induction, the given result is true for all natural numbers.

Question 2: For all n ≥ 1, prove that, 1.2.3 + 2.3.4 + 3.4.5+...+n(n + 1) (n + 2) = {n (n + 1) (n + 2) ( n + 3)} / 4

Solution: 

Let the given statement be S(n),

S(n):1.2.3+ 2.3.4 + 3.4.5+\ldots+ n.(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n+3)}{4}\\ \ \\ \text{For n=1,} \\ S(1):\frac{1(1+1)(1+2)(1+3)}{4} = 6\\ \text{which is true.}

Now, let's take a positive integer, k, and assume S(k) to be true i.e.
S(k):1.2.3+ 2.3.4 + 3.4.5+\ldots+ k.(k+1)(k+2) = \frac{k(k+ 1)(k + 2)(k+3)}{4}

We shall now prove that  S(k + 1) is also true, so now we have,
S(k+1):S(k) + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)}{4} + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)+ \ 4(k+1)(k+2)(k+3)}{4} \\ \ \\ \Rightarrow S(k+1):  \frac{(k+1)(k+2)(k+3)(k+4)}{4}\\ \ \\ \Rightarrow S(k+1): \frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\}\{(k+1)+3\} }{4} \\ 

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Question 3: For all n ≥ 1, prove that, 1 + 3 + 5 +... + 2n - 1 = n²

Solution: 

Let the given statement be S(n), 
and S(n) = 1 + 3 + 5 +... + 2n - 1 = n²
For n = 1,
L.H.S = 2 (1) × 1 - 1 = 1
R.H.S = 1² = 1

Thus S(1) is true .

Now, let's take a positive integer, k, and assume S(k) to be true i.e.,
S(k) = 1+ 3 + 5+...+(2k - 1) = k² 

We shall now prove that  S(k + 1) is also true, so now we have,
1 + 3 + 5+...+ (2(k + 1) - 1) = (k + 1)² 

L.H.S = 1 + 3 + 5 + .... (2k - 1 ) + 2k + 2 - 1

⇒ L.H.S = S(k) + 2k + 1
⇒ L.H.S = k² + 2k + 1
⇒ L.H.S = (k + 1)² 
⇒ L.H.S = R.H.S

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Question 4: For all n ≥ 1, prove that, 1.2 + 2.3 + 3.4 +...+ n(n + 1) = {n(n + 1)(n + 2)} / 3

Solution: 

Let the given statement be S(n),
\ S(n):1.2+ 2.3 + 3.4+ ……+ n.(n+1) = \frac{n(n + 1)(n + 2)}{3}\\ \ \\ \text{for n=1,} \\ \ S(1) : \frac{1(1+1)(1+2)}{3} = 2\\ \text{which is true.}

Now, let's take a positive integer, k, and assume S(k) to be true i.e.,
S(k):1.2+ 2.3 + 3.4+ ……+ k.(k+1) = \frac{k(k+ 1)(k + 2)}{3} \ \\       

We shall now prove that S(k + 1) is also true, so now we have,
S(k+1) : S(k) + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) : \frac{k(k+ 1)(k + 2)}{3} + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) :\frac{k(k+ 1)(k + 2)+ 3(k+1)(k+2)}{3} \\ \ \\ \Rightarrow S(k+1) :\frac{(k+1)(k+2)(k+3)}{3}\\ \ \\ \Rightarrow S(k+1) :\frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\} }{3}        

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Question 5: Prove that a_n = a₁ + (n - 1) d is the general term of any arithmetic sequence.

Solution: 

For n = 1,
a_n = a₁ + (1 - 1) d = a₁
so the formula holds true for n = 1

Let us assume that the formula a_k = a₁ + (k - 1) is true for all natural numbers. 

We shall now prove that the formula is also true for k+1, so now we have,
a_{k + 1} = a₁ + [(k + 1) - 1] d = a₁ + k · d.

We assumed that a_k = a₁ + (k - 1) d, and by the definition of an arithmetic sequence a_{k+ 1} - a_k = d,

Then, a_{k + 1} - a_k 
= (a₁ + k · d) - (a1 + (k - 1)d)
= a₁ - a₁ + kd - kd + d
= d

Thus the formula is true for k + 1, whenever it is true for k. And we initially showed that the formula is true for n = 1. Thus the formula is true for all natural numbers.

Practice Questions

Question 1: Prove that for all natural numbers n, the following holds: 1 + 2 + 3 + . . . + n = [n(n + 1)]/2.

Question 2: Show that for all integers n ≥ 1 : 1³ + 2³ + 3³ + . . . + n³ = [{n(n + 1)}/2]².

Question 3: Verify that for all natural numbers n: 2ⁿ > n².

Question 4: Prove that for all integers n ≥ 1: 1×2 + 2×3 + 3×4 + ⋯ + n(n+1) = [n(n+1)(n+2)]/3.

Question 5: Show that for all integers n ≥ 1: 7ⁿ − 4ⁿ is divisible by 3.