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

\(P(n): \dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...\)

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

For n=1,

L.H.S. = \( \dfrac{1}{3.5}\)

R.H.S.= \( \dfrac{1}{3(2.1+3)}=\dfrac{1}{3.5}\)

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

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

\( P(k): \dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...\)

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

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

Consider

\([\dfrac{1}{3.5}+ \dfrac{1}{5.7}+ \dfrac{1}{7.9}+...\)

\( + \dfrac{1}{(2k+1)(2k+3)}]+\)

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

\( =\dfrac{k}{3(2k+3)}+ \dfrac{1}{(2k+3)(2k+5)}\)......... [Using (i)]

= \( \dfrac{1}{(2k+3)}[\dfrac{k}{3}+\dfrac{1}{(2k+5)}]\)

= \( \dfrac{1}{(2k+3)}[\dfrac{k(2k+5)+3}{3(2k+5)}]\)

= \( \dfrac{1}{(2k+3)}[\dfrac{2k^2+5k+3}{3(2k+5)}]\)

= \( \dfrac{1}{(2k+3)}[\dfrac{2k^2+2k+3k+3}{3(2k+5)}]\)

= \( \dfrac{1}{(2k+3)}[\dfrac{2k(k+1)+3(k+1)}{3(2k+5)}]\)

= \( \dfrac{(k+1)(2k+3)}{3(2k+3)(2k+5)}\)

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

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 Abhisek | 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]

7^{2n} + 2^{3n – 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.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N