Let a _ n be n ^ th term of the series 5 + 8 + 14 + 23 + 35 + 50 +...

The given series is ${a}_{n}$, then we can write as

${a}_{n} = 5 + 8 + 14 + 23 + . . . . . . . . . . . . + {a}_{n}{a}_{n} = 5 + 8 + 14 + 23 + . . . . . . + {a}_{n - 1} + {a}_{n}$

Subtracting above equations, we get

$0 = 5 + 3 + 6 + 9 + . . . . + \left({a}_{n} - {a}_{n - 1}\right) - {a}_{n}$

$⇒ {a}_{n} = 5 + \left[\dfrac{\left(n - 1\right)}{2}\left(2 \times 3 + \left(n - 1 - 1\right)3\right)\right]$

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

$⇒ {a}_{n} = \dfrac{1}{2}\left(3{n}^{2} - 3n + 10\right)$

So,

$⇒ {a}_{40} = \dfrac{1}{2}\left(3 \times {40}^{2} - 3 \times 40 + 10\right)$

$⇒ {a}_{40} = \dfrac{1}{2}\left(4800 - 120 + 10\right)$

$⇒ {a}_{40} = 2345$

$⇒ {S}_{30} = \mathop{∑}_{n = 1}^{30}{a}_{n}$

$⇒ {S}_{30} = \dfrac{1}{2}\left(3\mathop{∑}_{n = 1}^{30}\left({n}^{2}\right) - 3\mathop{∑}_{n = 1}^{30}\left(n\right) + \mathop{∑}_{n = 1}^{30}10\right)$

$= \dfrac{1}{2}\left(\dfrac{3\left(30\right)\left(31\right)\left(61\right)}{6} - \dfrac{3\left(30\right)\left(31\right)}{2} + 10 \times 30\right)$

$⇒ {S}_{30} = 13635$

$⇒ {S}_{30} - {a}_{40} = 13635 - 2345 = 11290$

Therefore, the required value is $11290$.