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 | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.
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72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N
n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.