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

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

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

For n=1,

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

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

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

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

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

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

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

Consider

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

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

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

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

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

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

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

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

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

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 | 2 years 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