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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}+\cdots

is equal to 225 k, then k is equal to : 

  • Option 1)

     

    9

  • Option 2)

     

    27

  • Option 3)

     

    108

  • Option 4)

     

    54

Answers (1)

best_answer

 

Summation of series of natural numbers -

\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )
 

- wherein

Sum of first n natural numbers

1+2+3+4+------+n= \frac{n(n+1)}{2}

 

 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{3}=\left ( \sum_{k=1}^{n}K \right )^{2}

- wherein

1^{3}+2^{3}+3^{3}+-------n^{3}\\= \left ( 1+2+3+-----+n \right )^{2}

 

S=\sum_{r=1}^{15}\left ( \frac{3r}{4} \right )^{3}=\frac{27}{64}\sum r^{3}

=\frac{27}{64}\times \frac{\left ( 15\times 16 \right )^{2}}{4}

=27\times 225

K=27

 

 

 


Option 1)

 

9

Option 2)

 

27

Option 3)

 

108

Option 4)

 

54

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