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`

Answer

N/a

Explanation

Let `T_(n) = "nth term of " {1 xx 2^(2) + 2 xx 3^(2) + 3 xx 4^(2) + .... + n xx (n + 1)^(2)}`
` = n xx (n - 1)^(2)`
`= n xx (n^(2) + 2n + 1)`
`= n^(3) + 2n^(2) + n`
`rArr` `S_(n) = sumn^(3) + 2sum n^(2) + sum 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)`
Let `t_(n) = "nth term of the series " {1^(2) xx 2 + 2^(2) xx 3 + 3^(2) xx 4 + .... + n^(2) xx (n + 1)}`
` = n^(2) xx (n + 1) = n^(3) + n^(2)`
`rArr S_(n) = sum n^(3) + sum 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)`
`(1 xx 2^(2) + 2 xx 3^(2) + 3 xx 4^(2) + .... + n xx (n + 1)^(2))/(1^(2) xx2 + 2^(2) xx 3 + 3^(2) xx 4 + ... + 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." "` Hence Proved.