Question

Find the value of \(\sqrt {\left( {1 + 3 + 5 \cdots \cdots + 93} \right)\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right) \cdots \left( {1 - \frac{1}{{2209}}} \right)} \).
1. 3
2. 4
3. 2
4. \(\surd 2\)

Answer 1

Correct Answer - Option 4 : \(\surd 2\)

Given:

√ (1 + 3 + 5 + ....... + 93) ( 1 - (1/3) ) ( 1 - (1/4) ) ( 1 - (1/5) ) ....... ( 1 - (1/2209) ) 

Calculation:

 Let a = 1 + 3 + 5 + ........ + 93

b = ( 1 - (1/3) ) ( 1 - (1/4) ) ( 1 - (1/5) ) ....... ( 1 - (1/2209) )

From the above, we have, √ ab

a is in arithmetic progression. 

a_n = a + ( n - 1 )  d 

93 = 1 + ( n - 1 ) × 2          ∵ a_n = 93; a = 1; d = 2 ( 3 -1 = 2)

93 - 1 = ( n- 1 ) × 2

n = ( 92 / 2 ) + 1

n = 47

S_n = n / 2 [ 2a + ( n - 1 ) d ]

S₉₃ = 47 / 2 [ 2 × 1 + ( 47 - 1 ) × 2 ]

= 47 / 2 [ 94 ]

= 2209

S_n = a = 2209

Now,

b = ( 1 - (1/3) ) ( 1 - (1/4) ) ( 1 - (1/5) ) ....... ( 1 - (1/2209) ) 

= ( ( 2/3) ( 3/4) (4/5) .......... ( 2208/2209)

= 2 / 2209                           ∵ all the other numbers will be striked off 

√ ab = √( 2209)(2/2209)

= √ 2

∴ √ (1 + 3 + 5 + ....... + 93) ( 1 - (1/3) ) ( 1 - (1/4) ) ( 1 - (1/5) ) ....... ( 1 - (1/2209) )  = √ 2