The sum of the series
`1 + (1)/(3^(2)) + (1 *4)/(1*2) (1)/(3^(4))+( 1 * 4 * 7)/(1 *2*3)(1)/(3^(6)) + ..., ` is (a) `((3)/(2))^((1)/(3))` (b) `((5)/(4))^((1)/(3))` (c) `((3)/(2))^((1)/(6))` (d) None of these

Options

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

Answer

b

Explanation

Comparing the given series with
`1 + nx = (n(n-1))/(2!) x^(2) + ...,` we get nx` = (1)/(9)` ….(i)
and , `(n(n-1))/(2) x^(2) -(4)/(162)` …(ii)
Dividing (ii) by the square of (i), we get
`(n-1)/(2n) = 2 rArr n-1 = 4 n rArr n = - 1//3`
Puttiing ` n = - 1//3` in (i), we get `x = - 1//3`
`therefore ` Sum of the series = `(1 - (1)/(3))^(1//3) = ((2)/(3)^(-1//3) = ((3)/(2))`