The value of ∑ r = 2 100 r ( r + 1 ) ( r + 2 ) 2 r ( 2 − r ) is equal to

2 − r = ( r + 2 ) − 2 r , so r ( r + 1 ) ( r + 2 ) 2 r ( 2 − r ) = r ( r + 1 ) 2 r − ( r + 1 ) ( r + 2 ) 2 r + 1 . The sum telescopes: 2 ⋅ 3 2 2 − 101 ⋅ 102 2 101 = 3 2 − 101 ⋅ 102 2 101 . Also 101 ⋅ 102 2 101 = 101 ⋅ 51 2 100 , so (A) is the same value.

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediummultiple choice

The value of $\sum_{r = 2}^{100} \frac{2 ^{r} ( 2 - r )}{r ( r + 1 ) ( r + 2 )}$ is equal to

Options:(select one or more)

Answer:

A, B

Solution:

$2 - r = (r + 2) - 2 r$ , so $\frac{2 ^{r} ( 2 - r )}{r ( r + 1 ) ( r + 2 )} = \frac{2 ^{r}}{r ( r + 1 )} - \frac{2 ^{r + 1}}{( r + 1 ) ( r + 2 )}$ . The sum telescopes: $\frac{2 ^{2}}{2 \cdot 3} - \frac{2 ^{101}}{101 \cdot 102} = \frac{2}{3} - \frac{2 ^{101}}{101 \cdot 102}$ . Also $\frac{2 ^{101}}{101 \cdot 102} = \frac{2 ^{100}}{101 \cdot 51}$ , so (A) is the same value.

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

⏱ 2mℹ️ Source: QB

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