JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum of the first \(15\) terms of the series \({\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + .....\) is equal to \(225\,k,\) then \(k\) is equal to
- A \(108\)
- B \(27\)
- C \(54\)
- D \(9\)
Answer & Solution
Correct Answer
(B) \(27\)
Step-by-step Solution
Detailed explanation
\(S = {\left( {\frac{3}{4}} \right)^3} + {\left( {\frac{6}{4}} \right)^3} + {\left( {\frac{9}{4}} \right)^3} + {\left( {\frac{{12}}{4}} \right)^3} + ........15\,term\) \( = \frac{{27}}{{64}}\sum\limits_{r = 1}^{15} {{r^3}} \)…
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