`(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3n-1)(3n+2))=(n)/((6n+4))`

Explanation

माना
`P(n) : (1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3n-1)(3n+2))=(n)/((6n+4))`
यदि n = 1 हो तो
L.H.S. = `(1)/(2.5)=(1)/(10)`
तथा R.H.S. = `(1)/(6.1+4)=(1)/(6+4)=(1)/(10)`
implies L.H.S. = R.H.S.
अतः दिया हुआ कथन P(n), n = 1 के लिए सत्य है।
तब माना कि कथन P(n), n = k के लिए भी सत्य है।
`therefore P(k)=(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3k-1)(3k+2))=(k)/(6k+4)` n = k + 1 के लिए,
`P(k+1):(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/([(3k+1)-1][3(k+1)+2])`
`=(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3k-1)(3k+2))+(1)/((3k+2)(3k+5))`
`=((1)/(2.5)+(1)/(5.8)+(1)/(8.11)+...+(1)/((3k-1)(3k+2)))+(1)/((3k+2)(3k+5))`
`=(k)/(6k+4)+(1)/((3k+2)(3k+5))`
`=(1)/((3k+2))((k)/(2)+(1)/(3k+5))`
`=(1)/(3k+2)[(3k^(2)+5k+2)/(6k+10)]`
`=(1)/(3k+2)[(3k^(2)+3k+2k+2)/(6k+10)]`
`=(1)/(3k+2).[(3k(k+1)+2(k+1))/(6k+10)]`
`=(1)/(3k+2).((3k+2)(k+1))/(6k+10)=(k+1)/(6k+10)`
तब दिया हुआ कथन P(n), n = k + 1 के लिए भी सत्य है।
अतः दिया हुआ कथन P(n), n के प्रत्येक मान के लिए सत्य है जबकि `n in N`