If T_n denotes the nth term of the series $2+3+6+11+18+\dots$ then T₅₀ is 

Let, $\mathrm{S}=2+3+6+11+18+\dots +{\mathrm{T}}_{\mathrm{n}}$

or $\mathrm{S}=2+3+6+11+\dots +{\mathrm{T}}_{\mathrm{n}}$

On subtracting,  we get

 $0=2+1+3+5+7+\dots -{\mathrm{T}}_{\mathrm{n}}$

$\Rightarrow {\mathrm{T}}_{\mathrm{n}}=2+[1+3+5+\dots (\mathrm{n}-1\left)\text{ term }\right]$

$\begin{array}{l}=2+\frac{\mathrm{n}-1}{2}[2+(\mathrm{n}-1-1\left)2\right]\\ \begin{array}{rl}{\mathrm{F}}_{\mathrm{n}}& =2+(\mathrm{n}-1{)}^2\\ \therefore {\mathrm{T}}_{50}& =2+(50-1{)}^2={49}^2+2\end{array}\end{array}$