Let us denote the given equality by P(n), i.e.,
\( (1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....\)
\( (1+\dfrac{(2n+1)}{n^2})=(n+1)^2\)
For n=1,
L.H.S.= \( (1+ \dfrac{3}{1})=4\)
R.H.S.= (1+1)2 = 22 = 4
Therefore, P(n) is true for n=1.
Let us assume that P(k) is true for some positive integer k , i.e.,
\( (1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....\)
\( (1+\dfrac{(2k+1)}{k^2})=(k+1)^2\) ........(i)
Now, we have to prove that P( k+1) is also true.
Consider
\( [ (1+ \dfrac{3}{1})(1+\dfrac{5}{4})(1+\dfrac{7}{9})....\)
\( (1+\dfrac{(2k+1)}{k^2})]=\{1+\dfrac {\{2(k+1)+1\}}{(k+1)^2}\}\)
\( (k+1)^2(1+\dfrac {2(k+1)+1}{(k+1)^2})\)
\( (k+1)^2[\dfrac {(k+1)^2+2(k+1)+1}{(k+1)^2}]\)
\( (k+1)^2+2(k+1)+1\)
\( \{(k+1)+1\}^2\)
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.
a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]
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.