सिध्द कीजिए कि `(1xx2^(2)+2xx3^(2)+..+n(n+1)^(2))/(1^(2)xx2+2^(2)xx3+...+n^(2)(n+1))=(3n+5)/(3n+1)`

Explanation

माना अंश (Numerator)का nवाँ पद `T_(n)`हैं |
`:. " "T_(n)=n(n+1)^(2)=n^(3)+2n^(2)+n`
`:. " "अंश =SigmaT_(n)=Sigma(n^(3)+2n^(2)+n)=Sigman^(3)+2Sigma n^(2)+Sigma n`
`=(n^(2)(n+1)^(2))/(4)+2((n(n+1)(2n+1))/(6))+(n(n+1))/(2)`
` =(n(n+1))/(12)[3n(n+1)+4(2n+1)+6]`
`=(n(n+1)(3n^(2)+11n+10))/(12)`
`=(n(n+1)(n+2)(3n+5))/(12)" "`...(i)
इस प्रकार हर (denominator)का nवाँ पद `=n^(2)(n+1)=n^(3)+n^(2)`
`:. " " हर =Sigma n^(3)+Sigma n^(2)`
`=(n^(2)(n+1)^(2))/(4)+(n(n+1)(2n+1))/(6)=(n(n+1))/(12)[3n(n+1)+2(2n+1)]`
`=(n(n+1)(3n^(2)+7n+2))/(12)`
`=(n(n+1)(n+2)(3n+1))/(12)" "`...(ii)
समीकरण (i) व (ii)से
`(1xx2^(2)+2xx3^(2)+...+n(n+1)^(2))/(1^(2)xx2+2^(2)xx3+..+n^(2)(n+1))=(3n+5)/(3n+1)"
"`यही सिध्द करना था |