Let S r = 2 + 4 + 6 + … ( r terms). Then

S r = r ( r + 1 ) . Option A: ∑ r ( r + 1 ) 1 = 1 − n + 1 1 → 1 . True. Option B: ∑ S r = 6 n ( n + 1 ) ( 2 n + 1 ) + 2 n ( n + 1 ) = 3 n ( n + 1 ) ( n + 2 ) . So the option is false. Option C: ∑ k = 1 n k ( k + 1 ) ( k + 2 ) 3 = 2 3 ( 1 ⋅ 2 1 − ( n + 1 ) ( n + 2 ) 1 ) → 4 3 . True; option D is therefore false.

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediummultiple choice

Let $S_{r} = 2 + 4 + 6 + \dots$ ( $r$ terms). Then

Options:(select one or more)

Answer:

A, C

Solution:

$S_{r} = r (r + 1)$ .

Option A: $\sum \frac{1}{r ( r + 1 )} = 1 - \frac{1}{n + 1} \to 1$ . True.

Option B: $\sum S_{r} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} = \frac{n ( n + 1 ) ( n + 2 )}{3}$ . So the option is false.

Option C: $\sum_{k = 1}^{n} \frac{3}{k ( k + 1 ) ( k + 2 )} = \frac{3}{2} (\frac{1}{1 \cdot 2} - \frac{1}{( n + 1 ) ( n + 2 )}) \to \frac{3}{4}$ . True; option D is therefore false.

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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