MCQ
${n^{th}}$ term of the series $2 + 4 + 7 + 11 + .......$ will be
- A$\frac{{{n^2} + n + 1}}{2}$
- B${n^2} + n + 2$
- ✓$\frac{{{n^2} + n + 2}}{2}$
- D$\frac{{{n^2} + 2n + 2}}{2}$
$S = 2 + 4 + 7 + 11 + 16 + ........{T_{n - 1}} + {T_n}$
Subtracting, we get
$0 = 2 + \left\{ {2 + 3 + 4 + ........ + ({T_n} - {T_{n - 1}})} \right\} - {T_n}$
$ \Rightarrow $${T_n} = 1 + (1 + 2 + 3 + 4 + ......{\rm{upto}}\;n\;{\rm{terms}})$
$ \Rightarrow $$1 + \frac{1}{2}n(n + 1) = \frac{{2 + {n^2} + n}}{2} $
$= \frac{{{n^2} + n + 2}}{2}$.
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