Prove the following by using the principle of mathematical induction for all \( n\in N\)

\( (1+ \dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})....(1+\dfrac{1}{n})=(n+1)\)

Asked by Pragya Singh | 2 years ago |  86

1 Answer

Solution :-

Let us denote the given equality by P(n), i.e.,

P(n): \( (1+ \dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})....(1+\dfrac{1}{n})=(n+1)\)

For n=1,

L.H.S.= \( (1+ \dfrac{1}{1})=2\)

R.H.S.=(1+1) = 2

Therefore, P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,

\( (1+ \dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})....\)

\( (1+\dfrac{1}{k})=(k+1)\) .........(i)

Now, we have to prove that P(k+1) is also true.

Consider

\([P(k): (1+ \dfrac{1}{1})(1+\dfrac{1}{2})(1+\dfrac{1}{3})....\)

\( (1+\dfrac{1}{k})] (1+\dfrac{1}{k+1})\)

\( (k+1)(1+\dfrac{1}{k+1})\) .............[Using (i)]

\( (k+1)[\dfrac{(k+1)+1}{(k+1)}]\)

=(k+1)+1

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 2 years ago

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