If the sum of the first 15 terms of the series 3 4 3 + 1 1 2 3 + 2 1 4 3 + 3 3 + 3 3 4 3 + … is equal to 225 K , then K is equal to :

Given series, 3 4 3 + 6 4 3 + 9 4 3 + 12 4 3 + 15 4 3 + … 15 terms = 3 4 3 1 3 + 2 3 + 3 3 + … + 15 3 = 27 64 15 × 16 2 2 ∵ ∑ n = 1 n = n n 3 = n n + 1 2 2 = 27 64 × 225 × 64 1 = 27 × 225 Since, given 225 K = 27 × 225 ⇒ K = 27

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minPYQ_2019

MathematicsMediumstatement

If the sum of the first $15$ terms of the series ${(\frac{3}{4})}^{3} + {(1 \frac{1}{2})}^{3} + {(2 \frac{1}{4})}^{3} + 3^{3} + {(3 \frac{3}{4})}_{}^{3} + \dots$ is equal to $225 K$ , then $K$ is equal to :

Options:

Answer:

Solution:

Given series, ${(\frac{3}{4})}^{3} + {(\frac{6}{4})}^{3} + {(\frac{9}{4})}^{3} + {(\frac{12}{4})}^{3} + {(\frac{15}{4})}^{3} + \dots 15$ terms

$= {(\frac{3}{4})}^{3} (1^{3} + 2^{3} + 3^{3} + \dots + 15^{3})$

$= \frac{27}{64} {(\frac{15 \times 16}{2})}^{2} [∵ \sum_{n = 1}^{n = n} n^{3} = {[\frac{n (n + 1)}{2}]}^{2}]$

$= \frac{27}{64} \times \frac{225 \times 64}{1} = 27 \times 225$

Since, given $225 K = 27 \times 225$

$\Rightarrow K = 27$

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

⏱ 2mℹ️ Source: PYQ_2019

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