The sum of the series
` (3)/(4.8) - (3.5)/(4.8.12) + (3.5.7)/(4.8.12.16) - `… is equal to

Options

A. `sqrt((3)/(2)) - (3)/(4) `
B. `sqrt((2)/(3)) - (3)/(4) `
C. `sqrt((3)/(2)) -(1)/(4) `
D. `sqrt((2)/(3)) - (1)/(4) `

Answer

B

Explanation

Given tha,
` (3)/(4.8) - (3.5)/(4.8.12) + (3.5.7)/(4.8.12.16)- …. `
` = (3)/(2!4^2) - (3.5)/(4^3. 3!) + (3.5.7)/(4^4 4 ! ) - … `
` ( 1 + x ) ^n = 1 + (nx ) /(1!) + ( n (n - 1 ))/(2!) x ^2 + (n(n - 1 ) (n - 2))/(3!) x ^ 3 + (n(n -1) (n - 2 ) (n - 3 ))/(4!) x ^ 4 + ... `
On comparing the above expansion, we get
` therefore n(n -1 ) x^ 2 = (3)/(4^ 2 ) " " ...(2) `
`n ( n - 1 ) (n - 2 ) x ^3 = ( -3 (5))/(4^3)" " ...(3) `
` n(n - 1) (n - 2 ) (n - 3 ) x ^ 4 = ((3) (5)(7))/(4 ^ 4 ) " " `...(4)
` (3) div (2) , (4) div (3) `
` rArr (n - 2 ) x = ( -5 ) /(4) " " ...(5) `
` rArr (n - 3 ) x = ( -7) /(4 ) " " `...(6)
` (5) - (6) `
` x = ( -5 )/(4 ) + (7)/(4) `
` x = (1 )/(2) `
` (n-2 ) x = ( -5 )/(4 ) `
` (n -2 ) (1)/(2) = (-5)/(4) `
` (n - 2 ) = ( -5)/(2) `
` rArr n = ( - 1 ) /(2) `
` therefore (1 + x ) ^n = 1 + (nx ) /(1! ) + (n (n - 1 ))/(2! ) x ^ 2 + (n(n - 1 ) (n - 2 ))/( 3 ! ) x ^ 3 + ... `
`(1 + (1)/(2)) ^(- (1)/(2)) = 1 + ((-(1)/(2) )(1/2))/(1!) `
` + ((-(1)/(2))(-1/2 - 1))/(2!) ((1)/(2)) ^ 2 + ... `
` sqrt(2/3) = 1 - (1 )/(4) + (3 )/(4.8) - (3.5 )/(4.8.12) + (3.5.7)/(4.8.12.16) - ... `
` therefore (3)/(4.8) - (3.5)/(4.8.12) + (3.5.7)/(4.8.12.16) - ... = sqrt(2/3) - (3)/(4) `