 Mathematical Inductions Important question for CBSE Class 11 Maths

Prove the Following using Principle of Mathematical induction
1. Prove that for any positive integer number n , n 3 + 2 n is divisible by 3
2. Prove that
3.                  1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4
for all positive integers n.
4. For every n ∈ N, 2n3 + 3n2 + n is divisible by 6.
5. Prove by induction that
6. 1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) = n(n + 1)(n + 2)/ 3
7.  For every  n ≥ 2 , n3-n is multiple of 6
8. For every n ≥ 7, 3n ≥  n!
9.  For all n  ≥  1
10. (1+x)n ≥ 1+nx
Where  (1+x) > 0
11. If n ∈ N, then 1·3+2·4+3·5+4·6+··· + n(n+2) = n(n+1)(2n+7) /6
12. Prove that 3  +32 +3 3 +34 +··· +3n = (3n+1 −3)/ 2 for every n ∈ N.
13. Prove that 1/ 1 + 1/ 4 + 1/ 9 +··· + 1/ n2 ≤ 2− 1/ n
14.  For all n > 1, 8n – 3n is divisible by 5.
Solution to Problem 1:
Let Statement P (n) is defined by
n 3 + 2 n is divisible by 3 Step 1: Basic Step
We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n
1 3 + 2(1) = 3
3 is divisible by 3 hence p (1) is true. STEP 2: Inductive Hypothesis
We now assume that p (k) is true
k 3 + 2 k is divisible by 3
is equivalent to
k 3 + 2 k = 3 B , where B is a positive integer. Step 3: Inductive Steps
We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms
(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3
= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]
= 3 B+ 3 [ k 2 + k + 1 ] = 3 [ B + k 2 + k + 1 ] Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
Solution to Problem 2:
Statement P (n) is defined by
1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4 Step 1: Basic Step
We first show that p (1) is true.
Left Side = 1 3 = 1
Right Side = 1 2 (1 + 1) 2 / 4 = 1 hence p (1) is true. STEP 2: Inductive Hypothesis
We now assume that p (k) is true
1 3 + 2 3 + 3 3 + ... + k 3 = k 2 (k + 1) 2 / 4 Step 3: Inductive Steps
add (k + 1) 3 to both sides
1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3
factor (k + 1) 2 on the right side
= (k + 1) 2 [ k 2 / 4 + (k + 1) ] set to common denominator and group
= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4
= (k + 1) 2 [ (k + 2) 2 ] / 4 We have started from the statement P(k) and have shown that
1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4 Which is the statement P(k + 1).
Solution to Problem 3:
Let P(n)
2n3 + 3n2 + n is divisible by 6
Step 1: Basic Step
P(1)  is just that 2 + 3 + 1 is divisible by 6, which is trivial.
STEP 2: Inductive Hypothesis
We now assume that P (k) is true
Ie.
2k3 + 3k2 + k is divisible by 6
Step 3: Inductive Steps
We have to prove P(k+1)
Now
2(k + 1)3 + 3(k + 1)2 + (k + 1)
= 2(k3 + 3k2 + 3k + 1) + 3(k 2 + 2k + 1) + (k + 1)
= (2k 3 + 3k 2 + k) + (6k2+ 6k + 2 + 6k + 3 + 1)
= (2k3 + 3k2 + k) + 6(k2 + 2k + 1)
The first term is divisible by 6 since P(k)  is true and the second term is a multiple of 6. Hence, the last quantity is divisible by 6
Solution to Problem 4:
Statement P (n) is defined by
1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) = n(n + 1)(n + 2)/ 3
Step 1: Basic Step
We first show that p (1) is true.
Left Side = 1.2 = 2
Right Side = 1 (1 + 1)(1+2) / 3 = 2 hence p (1) is true. STEP 2: Inductive Hypothesis
We now assume that p (k) is true
1 · 2 + 2 · 3 + 3 · 4 + · · · k · (k + 1) = k(k + 1)(k + 2)/ 3
Step 3: Inductive Steps
We have to prove P(k+1)
Now
1 · 2 + 2 · 3 + 3 · 4 + · · · (k +1)· [(k+1) + 1) = (k+1)[(k +1)+ 1][(k+1) + 2]/ 3
Taking LHS
1 · 2 + 2 · 3 + 3 · 4 + · · · (k +1)· [(k+1) + 1)
=1 · 2 + 2 · 3 + 3 · 4 + · ·k(k+1) + (k +1)· [(k+1) + 1)
= k(k + 1)(k + 2)/ 3  + (k +1)· [(k+1) + 1)
= k(k + 1)(k + 2)/ 3  + (k +1)· (k+2)
=(k+1)(k+2) [ k/3   +1]
=(k+1)(k+2)(k+3)/3
Which is the statement P(k + 1).

Solution to Problem 6:
Let Statement P (n) is defined by
for all n > 7,  n! >  3n   Step 1: Basic Step
Let n = 7

n! >  3n  7!= 5040
37= 2187
So p(7) is true
STEP 2: Inductive Hypothesis

We now assume that p (k) is true That is, k! >  3k    Step 3: Inductive Steps
Let n = k + 1.
Then:
(k+1)!   =(k+1)k!
>(k+1) 3k
Now k    > 7
So (k+1) >3
>3. 3k
>3k+1
Then P(n) holds for n = k + 1, and thus for all n > 7
Solution to Problem 7:
Let Statement P (n) is defined by
(1+x)n ≥ 1+nx
Where  (1+x) > 0
Step 1: Basic Step
Let n = 1
(1+x)n ≥ 1+nx
(1+x) ≥ 1+x
Which is true
So p(1) is true
STEP 2: Inductive Hypothesis

We now assume that p (k) is true (1+x)k ≥ 1+kx
Step 3: Inductive Steps
Let n = k + 1.
Then:
(1+x)k+1 ≥ 1+(k+1)x
Taking the LHS
(1+x)k+1 =(1+x)(1+x)k
Now from hypothesis we know that
(1+x)k ≥ 1+kx
Also  (1+x) > 0
So        (1+x)k+1 ≥(1+x)( 1+kx)
≥[1+kx2+ (k+1)x]
Now kx2  is a positive quantity so we can say that
≥[1+ (k+1)x]
Which is P(k+1)
Solution to Problem 11:
Let Statement P (n) is defined by
for all n > 1, 8n – 3n is divisible by 5.
Step 1: Basic Step
Let n = 1.

Then the expression 8n – 3n evaluates to 81 – 31 = 8 – 3 = 5, which is clearly divisible by 5.

STEP 2: Inductive Hypothesis

We now assume that p (k) is true That is, that 8k – 3k is divisible by 5.
Step 3: Inductive Steps
Let n = k + 1.
Then:

8k+1 – 3k+1 = 8k+1 – 3×8k + 3×8k – 3k+1

= 8k(8 – 3) + 3(8k – 3k) = 8k(5) + 3(8k – 3k)
The first term in 8k(5) + 3(8k – 3k) has 5 as a factor (explicitly), and the second term is divisible by 5 (by assumption). Since we can factor a 5 out of both terms, then the entire expression, 8k(5) + 3(8k – 3k) = 8k+1 – 3k+1, must be divisible by 5.
Then P(n) holds for n = k + 1, and thus for all n > 1.