Question

If $T_n$ denotes the $nth$ term of the series $2+3+6+11+18+....$, then $T_{50}$ is

• A. ${49}^2-1$
• B. ${49}^2$
• C. ${50}^2+1$
• D. ${49}^2+2$

Answer 1

The correct option is D ${49}^2+2$

$\begin{array}{c}2+3+6+11+\mathrm{18................}\\ \Rightarrow differenceofseriesareinA.P\\ 3-2=1,6-3=3,11-6=5,18-11=7\\ \Rightarrow 1,3,5,7areinA.P\\ \Rightarrow Sumof\left(n-1\right)termofseries\left(1,3,5,7\right)+2\\ \Rightarrow a=1,d=2\\ \Rightarrow S_{49}=\frac{49}{2}\left[2\times 1+\left(49-1\right)\times 2\right]=49\left(1+49-1\right)\\ \Rightarrow S_{49}={49}^2,Togetthe{T}_{50}termadd2+sum\\ ofthe1+3+5+7+..............,T_{50}=\left({49}^2+2\right)Ans.\end{array}$