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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.

 


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