Question

While doing proofs by mathematical induction , why do we assume ourselves that a statement P(n) is true for n = k , and then we use it to prove P(n) is true for n=k+1?
And at the end , when we prove that it is true for n=k+1 , why do we conclude that the statement P(n) is true for all n ( But we didn't proved it for n=k , we assumed it ourselves) ??

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Answer 1

Answer:

1. Question⇒   While doing proofs by mathematical induction , why do we assume ourselves that a statement P(n) is true for n = k , and then we use it to prove P(n) is true for n=k+1? And at the end , when we prove that it is true for n=k+1 , why do we conclude that the statement P(n) is true for all n ( But we didn't proved it for n=k , we assumed it ourselves) ??

Step-by-step explanation:

• Now we will try to understand induction proof from an example. First we take a property of sum of n natural numbers.
• 1 + 2 + 3 + ……. + n = n(n+1)2
• The above set of natural numbers is property P (n) which is simply a formula of sum of n natural numbers. By using induction proof technique we need to prove that this formula holds true for all natural numbers. As stated before the first step is base step P (1).
• For P (1),  
• LHS = 1
• RHS = 1(1+1)2 = 1.
• So, LHS = RHS.
• It is proved that P (1) is true.
• Now in second step by using induction hypothesis of mathematical induction we assume P (k) is true.
•         1 + 2 + 3 + ……. + k = k(k+1)2
• We need to prove P(k + 1) is true by using P (k) true.
• For P(k + 1),
• LHS = 1 + 2 + 3 + ……. + k + (k + 1)  
•       = k(k+1)2 + (k+1) ………by using induction hypothesis
•       = (k+1)(k+2)2
•       = (k+1)((k+1)+12 = RHS for P(k + 1)
• P(k + 1) is true, whenever P(k) is true.  
• Thus P(1) is true and P(k + 1) is true whenever p(k) is true.  
• Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer 2

Answer :-

Actually the given sentence is not truly correct .

As first we check the given statement if it is true for n = 1 or not ,, and if it's true then we go further and check it for n = k and then n = k + 1

Now let us suppose a example of a sentence to clarify that why we check it for n = k + 1 directly without checking it for n = k .

Suppose the statement is the

Prove that 1² + 2² + 3² + ....... n²

$= \dfrac{n(n+1)(2n +1)}{6}$

Now while start proving we will first check it for n = 1

$= \dfrac{1(1+1)(2(1) +1)}{6}$

$= \dfrac{1(2)(3)}{6}$

$= \dfrac{6}{6}$

$= 1$

So it's true for n = 1

Now as we can clearly see that if we replace n = k we will get the same equation , which is :-

1² + 2² + 3² + ...... k²

$= \dfrac{k(k+1)(2k +1)}{6}$

So we can reduce the step and check it for k + 1

1² + 2² + 3² ...... k² + (k + 1)²

$= \dfrac{k(k+1)(2k +1)}{6} + ( k + 1)^2$

$= \dfrac{k(k + 1)(2k + 1) + 6(k + 1)^2}{6}$

$= \dfrac{ (k + 1)\left( k(2k + 1) + 6(k + 1)\right) }{6}$

$= \dfrac{ (k + 1)\left( 2k^2 + k + 6k + 6 \right) }{6}$

$= \dfrac{(k + 1)(k +2)(2k + 3)}{6}$

Now as we clearly see that we will get same result by placing n = k + 1 hence Proved.

So by above we can say that we can assume that n = k will satisfy the given equation only if it is satisfied with n = 1