Let us check P (n) is true for n = 1

P (1): \(\dfrac{ 1}{2.5} = \dfrac{1}{6.1}+4\) = \(\dfrac {1}{10} =\dfrac {1}{10}\)

P (1) is true.

Now,

Let us check for P (k) is true, and have to prove that P (k + 1) is true.

P (k): \( \dfrac{1}{2.5} +\dfrac{ 1}{5.8} +\dfrac{ 1}{8.11 }+ … +\dfrac{ 1}{(3k-1) (3k+2)} = \dfrac{k}{(6k+4)}\)

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

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

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

P (k + 1) is true.

Hence proved by mathematical induction.

Answered by Aaryan | 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