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
                 1 ³ + 2 ³ + 3 ³ + ... + n ³ = n ² (n + 1) ² / 4
                       for all positive integers n.

3. For every n ∈ N, 2n³ + 3n² + n is divisible by 6.
4. Prove by induction that
1 · 2 + 2 · 3 + 3 · 4 + · · · n · (n + 1) = n(n + 1)(n + 2)/ 3

5.  For every  n ≥ 2 , n³-n is multiple of 6
6. For every n ≥ 7, 3ⁿ ≥  n!
7.  For all n  ≥  1
(1+x)ⁿ ≥ 1+nx
Where  (1+x) > 0

8. If n ∈ N, then 1·3+2·4+3·5+4·6+··· + n(n+2) = n(n+1)(2n+7) /6
9. Prove that 3  +3² +3 ³ +3⁴ +··· +3ⁿ = (3ⁿ⁺¹ −3)/ 2 for every n ∈ N.
10. Prove that 1/ 1 + 1/ 4 + 1/ 9 +··· + 1/ n² ≤ 2− 1/ n
11.  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! >  3^k    Step 3: Inductive Steps
              Let n = k + 1.
Then:
(k+1)!   =(k+1)k!
>(k+1) 3^k
Now k    > 7
So (k+1) >3
>3. 3^k
>3^k+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 8^k – 3^k is divisible by 5.
Step 3: Inductive Steps
              Let n = k + 1.
Then:

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

     = 8^k(8 – 3) + 3(8^k – 3^k) = 8^k(5) + 3(8^k – 3^k)
The first term in 8^k(5) + 3(8^k – 3^k) 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, 8^k(5) + 3(8^k – 3^k) = 8^k+1 – 3^k+1, must be divisible by 5.
Then P(n) holds for n = k + 1, and thus for all n > 1.

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