Prove the following by the principle of mathematical induction: `1/(1. 4)+1/(4. 7)+1/(7. 10)+....+1/((3n-2)(3n+1))=n/(3n+1)`

Explanation

`"Let " P(n) :(1)/(1.4)+(1)/(4.7)+(1)/(7.10)+`
`......+(1)/((3n-1)(3n+1))=(n)/((3n+1))`
for n=1
`L.H.S. =(1)/(1.4)=(1)/(4),R.H.S. =(1)/(3.1+1)=(1)/(4)`
`:. L.H.S. =R.H.S.`
`rArr` P(n) is true for n=1
Let P (n) be true for n=k.
`:. P(k): (1)/(1.4)+(1)/(4.7)+(1)/(7.10)+`
`......+(1)/((3k-2)(3k+1))=(k)/(3k+1)`
for n=K+1
`P(k+1): (1)/(1.4)+(1)/(4.7)+(1)/(7.10)+`
`....+(1)/((3k-2)(3k+1))+(1)/((3k+1)(3k+4))`
`=(k)/(3k+1)+(1)/((3k+1)(3K+4))=(k(3k+4)+1)/((3k+1)(3k+4))`
`(3k^(2)+4k+1)/((3k+1)(3K+4)) =((3k+1)(K+1))/((3k+1)(3K+4))=(K+1)/(3K+4)`
`rArr` P(n) is also true for n=K +1
Hence from the principle of mathematical induction P (n) is true for all natural numbers n