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

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 8. sequence and series1 Mark
MCQ
The odd numbers are divided as follows

$\begin{array}{*{20}{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1&3\end{array}$

$\begin{array}{*{20}{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5&7&9&{11}\end{array}$

$\begin{array}{*{20}{c}}{13}&{15}&{17}&{19}&{21}&{23}\\.&.&.&.&.&.\\.&.&.&.&.&.\\.&.&.&.&.&.\end{array}$

Then the sum of ${n^{th}}$ row is

  • A
    ${2^{n - 2}}[{2^n} + {2^{n - 1}} - 1]$
  • B
    $\frac{1}{2}(2n + 1)$
  • C
    $2n$
  • ✓
    $4{n^3}$
✓

Answer

Correct option: D.
$4{n^3}$
d
(d) The first row contains $2$ numbers, the second row $4$, the third row $6$ and so on ${n^{th}}$ row contains $2n$ numbers whose first term ${(n - 1)^2} + {n^2}$ and $d = 2$.

Hence sum of $2n$ terms is
$n$.

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