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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
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
    $\frac{n}{6}({n^2} - 3n + 8)$
✓

Answer

Correct option: B.
$\frac{n}{6}({n^2} + 3n + 8)$
b
(b) We have $S = 2 + 4 + 7 + 11 + 16 + ..... + {T_n}$

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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