` 1 + (2)/(4) + (2.5)/(4.8) + (2.5.8)/(4.8.12) + (2.5.8.11)/(4.8.12.16) + … ` is equal to

Options

A. ` 4^( -2//3 ) `
B. `3sqrt (16) `
C. `3 sqrt4 `
D. ` 4^(3//2 ) `

Answer

B

Explanation

` 1 + (2)/(4) + (2.5)/(4.8) + (2.5.8)/(4.8.12) + … `
` (1 + x )^n = 1 + (nx )/(1 ! ) + (n (n - 1 ))/(2 ! ) x ^ 2 + (n(n - 1 ) (n - 2 )) /(3! ) x ^ 3 + … `
On comparing the above two expansions with , we get,
` therefore (nx )/( 1! ) = (2 )/(4)" " `...(1)
` (n(n - 1 ))/(2 ! ) x ^ 2 = (2.5)/(4.8)" " `...(2)
` (n(n - 1 )(n - 2 ))/(3 ! ) x ^ 3 = (2.5.8)/(4.8.12) " " `...(3)
(2) ` div` (1)
` ((n-1))/(2) x = (5)/(8) rArr (n-1) n = (5 )/(4) " " `...(4)
` (3) div (2) `
` ((n-2) x )/(3) = (8)/(12) rArr (n-2 ) x = 2 " " ...(5) `
` (4) - (5) `
` x = (5)/(4) - 2 `
` x = (-3)/(4) `
` therefore ((n-1))/(2)((-3)/(2)) = (5)/(8) `
` therefore n = (-2)/(3) `
` therefore 1 + (2)/(4) + (2.5)/(4.8) + (2.5.8)/(4.8.12) + ... = (1 - (3)/(4)) ^(-2/3) `
` = ((1)/(4)) ^(-2/3) `
` = 4 ^(2//3 ) `
` = 3sqrt(16)`