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

Explanation

Taking L.H.S.

`(1xx2^2+2xx3^2+......+nxx(n+1)^2)/(1^2xx2+2^2xx3+......+n^2xx(n+1))`

We will solve denominator & numerator separately

Solving numerator
Let numerator be
`S_1=(1xx2^2+2xx3^2+......+nxx(n+1)^2)`

`n^(th)` term is `nxx(n+1)^2`
Let `a_n=n(n+1)^2`
` " " " =n(n^2+1+2n)`
` " " " =n^3+n+2n^2`

Now finding `S_1`

`S_1=sum_(n=1)^na_n`

` " " " " =sum_(n=1)^n(n^3+n+2n^2)`

` " " " " =sum_(n=1)^n(n^3)+sum_(n=1)^n n+sum_(n=1)^n2n^2`

` " " " " =sum_(n=1)^n(n^3)+2sum_(n=1)^n n^2+sum_(n=1)^n n`

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

` " " " " =(n(n+1))/2((n(n+1))/2+(2(2n+1))/3+1)`

` " " " " =(n(n+1))/2((3n(n+1)+2xx2(2n+1)+6)/6)`

` " " " " =(n(n+1))/(2xx6)(3n(n+1)+2xx2(2n+1)+6)`

` " " " " =(n(n+1))/(12)(3n^2+3n+8n+4+6)`

` " " " " =(n(n+1))/(12)(3n^2+11n+10)`

` " " " " =(n(n+1))/(12)(3n^2+5n+6n+10)`

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

Thus, `S_1 =(n(n+1))/(12)((3n+2)(3n+5))`

Now solving denominator
Let denominator be
`S_2=1^2xx2+2^2xx3+......+n^2xx(n+1)`
`n^(th)` term is `n^2(n+1)`
Let `b_n=n^2(n+1)`
`b_n=n^3+n^2`

Now, calculating `S_2`

`S_2=sum_(n=1)^nb_n`

` " " " " =sum_(n=1)^n (n^3+n^2)`

` " " " " =sum_(n=1)^n n^3+sum_(n=1)^n n^2`

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

` " " " " =(n(n+1))/2[(n(n+1))/2+((2n+1))/3)`

` " " " " =(n(n+1))/2((3n(n+1)+2(2n+1))/(6))`

` " " " " =(n(n+1))/(2xx6)(3n(n+1)+2(2n+1))`

` " " " " =(n(n+1))/(12)(3n^2+7n+2)`

` " " " " =(n(n+1))/(12)(3n^2+6n+n+2)`

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

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

Thus, `S_2 =(n(n+1)(n+2)(3n+1))/(12)`

Now, Taking L.H.S.

`(1xx2^2+2xx3^2+......+nxx(n+1)^2)/(1^2xx2+2^2xx3+......+n^2xx(n+1))=(S_1)/(S_2)`

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

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

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

` " " " " =(3n+5)/(3n+1)`

` " " " " =` R.H.S.

Hence, L.H.S. = R.H.S.

Hence Proved