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

Asked by Aaryan | 1 year ago |  171

1 Answer

Solution :-

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 ago

Related Questions

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

Class 11 Maths Principle of Mathematical Induction View Answer