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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

\((1+\frac{1}1)\)\((1+\frac{1}2)\)\((1+\frac{1}3)\)......\(\left\{1+\frac{1}{n}\right\}\) = (n + 1).

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Best answer

To Prove:

\((1+\frac{1}1)\)\((1+\frac{1}2)\) x \((1+\frac{1}3)\) x......x \(\left\{1+\frac{1}{n^1}\right\}\) = (n + 1)1

Let us prove this question by principle of mathematical induction (PMI)

Let P(n): \((1+\frac{1}1)\)\((1+\frac{1}2)\) x \((1+\frac{1}3)\) x......x \(\left\{1+\frac{1}{n^1}\right\}\) = (n + 1)1

For n = 1 

LHS = 1 + \(\frac{1}1\) = 2

RHS = (1 + 1)2 = 2

Hence, LHS = RHS 

P(n) is true for n = 1 

Assume P(k) is true

 \((1+\frac{1}1)\)\((1+\frac{1}2)\) x \((1+\frac{1}3)\) x......x \(\left\{1+\frac{1}{k^1}\right\}\) = (k + 1)1.....(1)

We will prove that P(k + 1) is true

[Now writing the second last term]

\((1+\frac{1}1)\)\((1+\frac{1}2)\) x \((1+\frac{1}3)\) x......x \(\left\{1+\frac{1}{k^1}\right\}\) x {1 + \(\frac{1}{(k+1)^1}\)}

 

LHS = RHS 

Therefore, P (k + 1) is true whenever P(k) is true. 

By the principle of mathematical induction, P(n) is true for 

Where n is a natural number 

Hence proved.

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