MCQ
$2 + 4 + 7 + 11 + 16 + ......$to $n$ terms =
- A$\frac{1}{6}({n^2} + 3n + 8)$
- ✓$\frac{n}{6}({n^2} + 3n + 8)$
- C$\frac{1}{6}({n^2} - 3n + 8)$
- D$\frac{n}{6}({n^2} - 3n + 8)$
Again $S = {\rm{ }}2 + 4 + 7 + 11 + ....... + {T_{n - 1}} + {T_n}$
Subtracting, we get
$0 = 2 + \left\{ {2 + 3 + 4 + 5 + .....({T_n} - {T_{n - 1}})} \right\} - {T_n}$
${T_n} = 2 + \frac{1}{2}(n - 1)(4 + \{ n - 2)1\} = \frac{1}{2}({n^2} + n + 2)$
Now $S = \Sigma {T_n} = \frac{1}{2}\Sigma ({n^2} + n + 2) $
$= \frac{1}{2}(\Sigma {n^2} + \Sigma n + 2\Sigma \,1)$
$ = \frac{1}{2}\left\{ {\frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{2}n(n + 1) + 2n} \right\}$
$ = \frac{n}{{12}}\left\{ {(n + 1)(2n + 1 + 3) + 12} \right\}$
= $\frac{n}{6}\left\{ {(n + 1)(n + 2) + 6} \right\} $
$= \frac{n}{6}({n^2} + 3n + 8)$.
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