Question

If $t_n$ denotes $n$th term of the series $2+3+6+11+\mathrm{18...}$, then $t_{52}=$

• A. $2+{49}^2$
• B. $2+{48}^2$
• C. $2+{50}^2$
• D. $2+{51}^2$

Answer 1

Answer: (D)

$2+{51}^2$

$S=2+3+6+11+\mathrm{18........}T\left(52\right)$
Subtracting the former term from the later like 2nd term-1st term .
$T\left(52\right)=2+1+3+5+....101$
**Now it should be kept in mind that sum of first n odd numbers is $n^2$**(Important)
Therefore $T\left(52\right)=2+(51{)}^2$.
Option D