Show that `(1xx2^2+2xx3^2+. . . +nxx(n+1)^2)/(1^2xx2+2^2xx3+. . . +n^2xx(n+1))=(3n+5)/(3n+1)dot`

Explanation

We have to proof `(1xx2^2+2xx3^2+ldots+nxx(n+1)^2)/(1^2xx2+2^2xx3+ldots+n^2xx(n+1))=(3n+5)/(3n+1)`

Then the `t^(th)` term will be

`=(sum_(r=1)^(n) (r^3+r^2+r))/((sum_(r=1)^(n) (r^3+r^2)) `

`=((n^2(n+1)^2)/4+(2n(n+1)(2n+1))/6+(n(n+1))/2)/((n^2(n+1)^2)/4+(n(n+1)(2n+1))/6)`



`=((n(n+1))/12[3n^2+11n+10])/((n(n+1))/12[3n^2+6n+n+2])`



`=((n(n+1)(n+2)(3n+5))/12)/((n(n+1)(n+2)(3n+1))/12)`

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