Question

If $T_n$ denotes the nth terms of the series.$2+3+6+11+18+..........$ , then $T_{5}0$ is :

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

Answer 1

Answer: (D)

${49}^2+2$

$2+3+6+11+18+.................$

$\begin{array}{c}{T}_n=a{x}^2+bx+c\\ T_1=a+b+c=2\\ T_2=a+b+c=3\\ T_3=a+b+c=6\\ 3a+b=1\\ 5a+b=3\\ 2a=2\\ a=1\\ b=-2\\ c=3\end{array}$

$T_n=n^2-2n+3$

$T_{50}={\left(50\right)}^2-2\left(50\right)+3$

$T_{50}={\left(49\right)}^2+2$

Hence,

Option $D$ is correct answer.