Let us check for n = 1,
P (1): \(\dfrac{ 1}{3.7} = \dfrac{1}{(4.1-1)(4+3)}\)
\(\dfrac{ 1}{21} = \dfrac{1}{21}\)
P (n) is true for n =1.
Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.
P (k): \(\dfrac{ 1}{3.7} + \dfrac{1}{7.11 }+ \dfrac{1}{11.15} + … + \dfrac{1}{(4k-1)(4k+3)} = \dfrac{k}{3(4k+3)}\) …. (i)
Substituting the value of P (k) we get,
=\(\dfrac{ k}{(4k+3) }+ \dfrac{1}{(4k+3)(4k+7)}\)
= \( \dfrac{\dfrac{1}{(4k+3) k(4k+7)+3}}{3(4k+7)}\)
=\(\dfrac{ (k+1) }{ 3(4k+7)}\)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.
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]
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.