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

\(P(n): \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+... \)

\( +\dfrac{1}{(3n-1)(3n+2)}=\dfrac{n}{(6n+4)}\)

For n=1,

L.H.S. = \( \dfrac{1}{2.5}=\dfrac{1}{10}\)

R.H.S.= \( \dfrac{1}{6.1+4}=\dfrac{1}{10}\)

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

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

\( \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+... \)

\( +\dfrac{1}{(3k+1)(3k+2)}=\dfrac{k}{(6k+4)}\).....(i)

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

Consider

\( \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+... \)

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

\(\dfrac{k}{6k+4}+ \dfrac{1}{(3k+3-1)(3k+3+2)}\)

\( \dfrac{k}{6k+4}+ \dfrac{1}{(3k+2)(3k+5)}\)

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

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

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

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

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

\( \dfrac{(k+1)}{6(k+1)+4}\)

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]

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