Question

1/1.4+1/4.7+1/7.10+......+1/(3n-2)(3n+1)=n/(3n+1)

Answer 1

L.H.S,   1/1.4+1/4.7+1/7.10+.............+1/(3n-2)(3n+1)
=1/3[3/1.4+3/4.7+3/7.10+........+3/(3n-2)(3n+1)]
=1/3[(1-1/4)+(1/4-1/7)+(1/7-1/10)+........+(1/(3n-2) - 1/(3n+1))]
Same numbers having positive and negative signs are canceled out except "1" and "-1/(3n+1)"
=1/3[1-1/(3n+1)]
=1/3[(3n+1)-1 /(3n+1)]
 =1/3[3n+1-1/(3n+1)]
  "1" and "-1" are canceled out
  =1/3[3n/(3n+1)]
  "3" in "1/3" and "3" in "3n" are canceled out
  =n/(3n+1)
  =R.H.S.
Hence Proved