Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

EXPLANATION.

$\implies \dfrac{1^{3} }{1} + \dfrac{1^{3} + 2^{3} }{1 + 3} + \dfrac{1^{3} + 2^{3} + 3^{2} }{1 + 3 + 5} + . . . . .$

Sum of first 9 terms,

As we know that,

We can write equation as,

$\implies \dfrac{1^{3} + 2^{3} + 3^{3} + . . . . . + n^{3} }{1 + 3 + 5 + . . . . . + n \ terms}$

1 + 3 + 5 + . . . . . + n terms.

First term = a = 1.

Common difference = d = b - a = c - b.

Common difference = d = 3 - 1 = 2.

Sum of n terms of an A.P.

⇒ Sₙ = n/2[2a + (n - 1)d].

⇒ Sₙ = n/2[2 + (n - 1)2].

⇒ Sₙ = n/2[2 + 2n - 2].

⇒ Sₙ = n/2[2n].

⇒ Sₙ = n².

As we know that,

Sum of cube of first n natural numbers.

⇒ ∑r³ = [n(n + 1)/2]².

Using this formula in the equation, we get.

$\implies \dfrac{\bigg(\dfrac{n(n + 1)}{2} \bigg)^{2} }{n^{2} } \ = \dfrac{(n + 1)^{2} }{4}$

$\implies \displaystyle \dfrac{1}{4} \sum_{n = 1}^{9} (n + 1)^{2}$

$\implies \displaystyle \dfrac{1}{4} \sum_{n = 1}^{9} (n^{2} + 2n + 1)$

$\implies \displaystyle \dfrac{1}{4} \bigg[\sum_{n = 1}^{9} n^{2} + \sum_{n= 1}^{9} 2n + \sum_{n = 1}^{9} 1 \bigg]$

$\implies \displaystyle \dfrac{1}{4} \bigg[\dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{n(n + 1)}{2} + n \bigg]$

$\implies \displaystyle \dfrac{1}{4} \bigg[\dfrac{9(9 + 1)(2(9) + 1)}{6} + \dfrac{9(2)((9) + 1)}{2} + 9 \bigg]$

$\implies \dfrac{1}{4} \bigg[ \dfrac{9 (10)(19)}{6} + 90 + 9 \bigg]$

$\implies \dfrac{1}{4} \bigg[ 3(5)(19) + 90 + 9 \bigg]$

$\implies \dfrac{1}{4} \bigg[ 285 + 90 + 9 \bigg]$

$\implies \dfrac{1}{4} \bigg[ 384 \bigg] = 96$

Option [B] is correct answer.

Answer 2

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

$\sf \: \dfrac{1^{3} }{1} + \dfrac{1^{3} + 2^{3} }{1 + 3} + \dfrac{1^{3} + 2^{3} + 3^{3} }{1 + 3 + 5} + . . . . .9 \: terms$

$\sf \: \dfrac{1^{3} }{1} + \dfrac{1^{3} + 2^{3} }{4} + \dfrac{1^{3} + 2^{3} + 3^{3} }{9} + . . . . .9 \: terms$

$\sf \: \dfrac{1^{3} }{ {1}^{2} } + \dfrac{1^{3} + 2^{3} }{ {2}^{2} } + \dfrac{1^{3} + 2^{3} + 3^{3} }{ {3}^{2} } + . . . . .9 \: terms$

From sequences we know,

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

$\sf \: 1^{2} + 2^{2} + 3^{2} + - - - + n^{2} = \bigg(\dfrac{n(n + 1)(2n + 1)}{6} \bigg)$

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

Now,

The nth term of the above sequence can be rewritten as

$\sf \: a_n \: = \: \dfrac{ \sum \: {n}^{3} }{ {n}^{2} }$

$\sf \: a_n \: = \: \dfrac{ \bigg(\dfrac{n(n + 1)}{2} \bigg) ^{2} }{ {n}^{2} }$

$\sf \: a_n = \dfrac{ {n}^{2} + 2n + 1}{4}$

So, Sum of 9 terms is given by

$\sf \:\sum_{n=1}^9 a_n =\sum_{n=1}^9 \dfrac{ {n}^{2} + 2n + 1}{4}$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}( \sum_{n=1}^9 {n}^{2} + 2\sum_{n=1}^9n + \sum_{n=1}^91)$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}\bigg(\dfrac{9(9 + 1)(18 + 1)}{6} + 9(9 + 1) + 9 \bigg)$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}\bigg(\dfrac{9(10)(19)}{6} + 9(10) + 9 \bigg)$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}\bigg((15)(19)+ 90 + 9 \bigg)$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}\bigg(285+ 90 + 9 \bigg)$

$\sf \:\sum_{n=1}^9 a_n =\dfrac{1}{4}\bigg(384 \bigg)$

$\sf \:\sum_{n=1}^9 a_n = 96$

Hence,

Option (b) is correct.