If S _ n = 4 + 11 + 21 + 34 + 50 + to n terms, then (1)/(60)( S _...

Given,

${S}_{n} = 4 + 11 + 21 + 34 + . . . . . . . . . . . . + {T}_{n}{S}_{n} = 4 + 11 + 21 + 34 . . . . . . + {T}_{n - 1} + {T}_{n}$

Subtracting above equations, we get

$0 = 4 + 7 + 10 + 13 + . . . . - {T}_{n}$

$⇒ {T}_{n} = 4 + 7 + 10 + 13 + . . .$

The above series is in AP.

$⇒ {T}_{n} = \dfrac{n}{2}\left(2 \times 4 + \left(n - 1\right)3\right)$

$⇒ {T}_{n} = \dfrac{n}{2}\left(3n + 5\right)$

So,

${S}_{n} = \mathop{∑}_{}^{}\dfrac{3{n}^{2} + 5n}{2}$

$⇒ {S}_{n} = \dfrac{1}{2}\left(\dfrac{3n ( n + 1 ) ( 2n + 1 )}{6} + \dfrac{5n\left(n + 1\right)}{2}\right)$

$⇒ {S}_{n} = \dfrac{n\left(n + 1\right)}{2}\left(\dfrac{( 2n + 1 )}{2} + \dfrac{5}{2}\right)$

$⇒ {S}_{29} = \dfrac{29 \times 30}{2}\left(\dfrac{59}{2} + \dfrac{5}{2}\right)$

$⇒ {S}_{29} = 29 \times 15 \times 32 = 13920$

$⇒ {S}_{9} = \dfrac{9 \times 10}{2}\left(\dfrac{19}{2} + \dfrac{5}{2}\right)$

$⇒ {S}_{9} = 9 \times 5 \times 12 = 540$

$⇒ \dfrac{{S}_{29} - {S}_{9}}{60} = \dfrac{13920 - 540}{60} = 223$.

Therefore, the required answer is $223$.