` 1 + (1)/(4) + (1.3)/(4.8) + (1.3.5)/(4.8.12) + `… is equal to

Options

A. `sqrt2`
B. `(1)/(sqrt2) `
C. `sqrt3 `
D. `(1)/(sqrt3)`

Answer

A

Explanation

` 1 + (1)/(4) + (1.3)/(4.8) + (1.3.5)/(4.8.12) + … `
` (1 + x ) ^n = 1 + nx + (n (n- 1 ))/(2!) x^ 2 `
` + (n(n - 1 )(n - 2 ))/(3!) x ^ 3 + … `
On comparing the above expansion, we get,
` therefore nx = (1)/(4) " "`...(1)
` (n(n - 1 ))/(2!) x ^ 2 = (1.3)/(4.8) " "`...(2)
` (n (n - 1 )(n -2))/(3!) x ^ 3 = (1.3.5 ) /(4.8.12)" " `...(3)
(2) `div` (1)
`((n-1))/(2) x = (3)/(2xx 4 ) `
` (n - 1 ) x = (3)/(4) " " `...(4)
From (3) ` div ` (2)
` ((n-2))/(3) x = (5)/(3) xx (1)/(4) " " `...(5)
From (4) and (5) , ` x = ( - 1 ) /(2) , n = ( - 1 ) /(2) `
` therefore 1 + (1)/(4) + (1.3)/(4.8) + (1.3.5 )/(4.8.12) + ... = (1 - (1)/(2) ) ^( -1/2) `
` = ((1)/(2))^( -1/2) = sqrt2 `