If the sum of the first 15 terms of the series [( 3 4 )^3 + ( 11 2 )^3 + ( 21 4 )^3 + 3^3 …

Question

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

Accepted Answer

S = $${\left( {{3 \over 4}} \right)^3} + {\left( {{6 \over 4}} \right)^3} + {\left( {{9 \over 4}} \right)^3} + {\left( {{{12} \over 4}} \right)^3} + \,........15$$ term

= $${{27} \over {64}}$$ $$\sum\limits_{r = 1}^{15} {{r^3}} $$

= $${{27} \over {64.}}{\left[ {{{15\left( {15 + 1} \right)} \over 2}} \right]^2}$$

= 225 K (Given in question)

K = 27

Solutions · 1