If the sum of the first 15 terms of the serie | vedclass

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(B) The given series is ${\left( {\frac{3}{4}} \right)^3} + {\left( {\frac{6}{4}} \right)^3} + {\left( {\frac{9}{4}} \right)^3} + {\left( {\frac{12}{4

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) The given series is ${\left( {\frac{3}{4}} ight)^3} + {\left( {\frac{6}{4}} ight)^3} + {\left( {\frac{9}{4}} ight)^3} + {\left( {\frac{12}{4}} ight)^3} + \dots$ up to $15$ terms.This can be written as $\sum_{r=1}^{15} {\left( \frac{3r}{4} ight)^3}$.Taking the constant out,we get $\left( \frac{3}{4} ight)^3 \sum_{r=1}^{15} r^3 = \frac{27}{64} \sum_{r=1}^{15} r^3$.Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} ight]^2$.For $n=15$,$\sum_{r=1}^{15} r^3 = \left[ \frac{15 	imes 16}{2} ight]^2 = (15 	imes 8)^2 = 120^2 = 14400$.Thus,the sum $S = \frac{27}{64} 	imes 14400$.$S = 27 	imes \frac{14400}{64} = 27 	imes 225$.Given that $S = 225k$,we have $225k = 27 	imes 225$.Therefore,$k = 27$.

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

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