दर्शाइए कि :
`(1xx2^(2)+2xx3^(2)+..+nxx(n+1)^(2))/(1^(2)xx2+2^(2)xx3+...+n^(2)xx(n+1))=(3n+5)/(3n+1)`

Explanation

माना `T_(n)=` श्रेणी `{1xx2^(2)+2^(2)+2xx3^(2)+3xx4^(2)+...+nxx(n+1)^(2)}` का n वां पद
`=nxx(n+1)^(2)`
`=n xx (n^(2)+2n+1)`
`=n^(3)+2n^(2)+n`
`rArr S_(n)=Sigman^(3)+2Sigma n^(2)+Sigma n`
`=(1)/(4)n^(2)(n+1)^(2)+(2)/(6)n(n+1)(2n+1)+(1)/(2)n(n+1)`
`=n(n+1)[(1)/(4)n(n+1)+(1)/(3)(2n+1)+(1)/(2)]`
`=n(n+1)[(3n(n+1)+4(2n+1)+6)/(12)]`
`=(1)/(12)n(n+1)(3n^(2)+3n+8n+4+6)`
`=(1)/(12)n(n+1)(3n^(2)+11n+10)`
`=(1)/(12)n(n+1)(n+2)(3n+5)`
माना `t_(n)=`श्रेणी `{1^(2)xx2+2^(2)xx3+3^(2)xx4+...+n^(2)xx(n+1)}` का n वां पद
`=n^(2)xx(n+1)=n^(3)+n^(2)`
`rArr s_(n)=Sigma n^(3)+Sigma n^(2)`
`=(1)/(4)n^(2)(n+1)^(2)+(1)/(6)n(n+1)(2n+1)`
`=n(n+1)[(n(n+1))/(4)+(2n+1)/(6)]`
`=n(n+1)[(3n(n+1)+2(2n+1))/(12)]`
`=(1)/(12)n(n+1)(3n^(2)+3n+4n+2)`
`=(1)/(12)n(n+1)(3n^(2)+7n+2)`
`=(1)/(12)n(n+1)(n+2)(3n+1)`
अब `L.H.S. =(1xx2^(2)+2xx3^(2)+3xx4^(2)+...+nxx(n+1)^(2))/(1^(2)xx2+2^(2)xx3+3^(2)xx4+...+n^(2)xx(n+1))`
`=(S_(n))/(s_(n))=((1)/(12)n(n+1)(n+2)(3n+5))/((1)/(12)n(n+1)(n+2)(3n+1))`
`=(3n+5)/(3n+1)=R.H.S.` यही सिद्ध करना था।