Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} +{\bigg( 1\dfrac{1}{2}\bigg) }^{3} + {\bigg( 2\dfrac{1}{4}\bigg) }^{3} + {3}^{3} + {\bigg( 3\dfrac{3}{4}\bigg) }^{3} ...$

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} +{\bigg( \dfrac{3}{2}\bigg) }^{3} + {\bigg( \dfrac{9}{4}\bigg) }^{3} + {3}^{3} + {\bigg( \dfrac{15}{4}\bigg) }^{3} ...$

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} +{\bigg(2 \times \dfrac{3}{4}\bigg) }^{3} + {\bigg(3 \times \dfrac{3}{4}\bigg) }^{3} + {\bigg( 4 \times \dfrac{3}{4} }\bigg)^{3} + {\bigg(5 \times \dfrac{3}{4}\bigg) }^{3} ...$

Taking out (3/4)³ as common :-

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} \bigg(1 + {2}^{3} + {3}^{3} + {4}^{3} + ...{15}^{3} \bigg)$

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} {\bigg( \dfrac{15(15 + 1)}{2} \bigg)}^{2}$

$\rm \longrightarrow {\bigg( \dfrac{3}{4}\bigg) }^{3} {\bigg( \dfrac{15 \times 16}{2} \bigg)}^{2}$

$\rm \longrightarrow {\bigg( \dfrac{27}{64}\bigg) } {\bigg( \dfrac{15 \times 8}{1} \bigg)}^{2}$

Sum of the series = 225k

$\rm \therefore {\bigg( \dfrac{27}{64}\bigg) } {\bigg( \dfrac{15 \times 8}{1} \bigg)}^{2} = 225k$

$\rm \longrightarrow \dfrac{27}{64} \times {(15)}^{2} \times {(8)}^{2} = 225k$

$\rm \longrightarrow \dfrac{27}{64} \times 225 \times 64 = 225k$

$\rm \longrightarrow \dfrac{27}{1} \times 225 \times 1 = 225k$

$\rm \longrightarrow \dfrac{27 \times 225}{225} = k$

$\rm \longrightarrow 27 = k$

Answer :- option (b)27 is correct

Answer 2

$\large\underline{\sf{Solution-}}$

It is given that

$\rm \: \bigg(\dfrac{3}{4} \bigg)^{3} + \bigg(\dfrac{3}{2} \bigg)^{3} + \bigg(\dfrac{9}{4} \bigg)^{3} + \bigg(\dfrac{3}{1} \bigg)^{3} + - - - 15 \: terms = 225k$

It can further written as

$\rm \: \bigg(\dfrac{3}{4} \bigg)^{3} + \bigg(\dfrac{2.3}{4} \bigg)^{3} + \bigg(\dfrac{3.3}{4} \bigg)^{3} + \bigg(\dfrac{3.4}{4} \bigg)^{3} + - - - 15 \: terms = 225k$

$\rm \: \bigg(\dfrac{3}{4} \bigg)^{3} \bigg \{1+ 2^{3} +3^{3} + 4^{3} + - - - 15 \: terms \bigg\} = 225k$

$\rm \: \bigg(\dfrac{27}{64} \bigg) \bigg \{ {1}^{3} + 2^{3} +3^{3} + 4^{3} + - - - 15 \: terms \bigg\} = 225k$

We know from sequences,

$\rm :\longmapsto\: {1}^{3} + {2}^{3} + {3}^{3} + - - - + {n}^{3} = \bigg(\dfrac{n(n + 1)}{2} \bigg)^{2}$

So, on applying this identity,

$\rm \: \dfrac{27}{64} \times \bigg(\dfrac{15(15 + 1)}{2} \bigg)^{2} = 225k$

$\rm \: \dfrac{27}{64} \times \bigg(\dfrac{15 \times 16}{2} \bigg)^{2} = 225k$

$\rm \: \dfrac{27}{64} \times \bigg(15 \times 8 \bigg)^{2} = 225k$

$\rm \: \dfrac{27}{64} \times \bigg(225 \times 64 \bigg) = 225k$

$\bf\implies \:k = 27$

Hence,

Option (b) is correct.