If t_n denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + ... then t₅₀ is

Given that: t_n be the nth term of the series

Let S_n = 2 + 3 + 6 + 11 + 18 +…+ t₅₀

Using method of difference, we get

S_n = 2 + 3 + 6 + 11 + 18 + … + t₅₀ …(i)

and S_n = 0 + 2 + 3 + 6 + 11 + … + t₄₉ + t₅₀ …(ii)

Subtracting eq. (ii) from eq. (i), we get

0 = 2 + 1 + 3 + 5 + 7 + … - t₅₀ terms

⇒ t₅₀ = 2 + (1 + 3 + 5 + 7 + … upto 49 terms)

We know that,

⇒ t₅₀ = 2 + 49 + 48 × 49

⇒ t₅₀ = 2 + 49(1 + 48)

⇒ t₅₀ = 2 + 49 × 49

⇒ t₅₀ = 2 + 49²

Hence, the correct option is (d)