Evaluate ∑ r = 1 10 r 2 ( 1 3 + 2 3 + ⋯ + r 3 ) ( 1 + 2 + ⋯ + r ) ( 1 + 3 + ⋯ + ( 2 r − 1 )) = a / b in lowest terms. Find a − b .

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

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Passage / Comprehension

Evaluate $\sum_{r = 1}^{10} \frac{( 1 + 2 + \dots + r ) ( 1 + 3 + \dots + ( 2 r - 1 ))}{r ^{2} ( 1 ^{3} + 2 ^{3} + \dots + r ^{3} )} = a / b$ in lowest terms. Find $a - b$ .

Options:

Answer:

9.00

Solution:

Using the standard sums, the $r$ th term reduces to $2/ [r (r + 1)] = 2 (1/ r - 1/ (r + 1))$ . The telescoping sum is $2 (1 - 1/11) =$ $20/11$ , so $a - b = 20 - 11 = 9$ .

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

⏱ 2mℹ️ Source: QB

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