If ` alpha = (5)/(2!3!) + (5.7)/(3!3^2) + (5.7.9)/(4!3^3) + …`, then ` alpha ^ 2 + 4 alpha ` is equal to

Options

A. 21
B. 23
C. 25
D. 27

Answer

B

Explanation

Given that
` alpha = (5 ) /(2!3 ) + (5.7 )/(3!3^ 2 ) + (5.7.9)/(4!3^3) +… `
` ( 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 + ... " " ` ...(2)
On comparing the terms of above two expansions, we get,
` (n(n - 1))/(2!)x ^ 2 = (5 )/(2!3)" " `...(1)
` (n(n - 1)(n - 2))/(3 ! ) x ^ 3 = (5.7 ) /(3 ! 3^ 2 ) " " `...(2)
` (n (n - 1 )(n - 2 ) (n - 3 ))/(4 ! ) x ^ 4 = (5.7.9)/(4!3^3) " " `...(3)
` (2) div (1) , (3) div (2) `
` (n - 2 ) x = (7 ) /(3) " " `...(4)
` (n - 3 ) x = 3 " " `... (5)
` (7)/(3x) - (3 )/(x) = 1 " " ` [From (4) and (5) ]
` ( -2 ) /(3x ) = 1 `
` rArr x = (-2)/(3 ) `
`rArr (n - 2 ) (( -3 )/(2)) = (7)/(2) `
` rArr n = 2 - (7)/(2)`
` rArr n = (-3)/(2) `
` therefore (1 - (2)/(3))^(-3/2) = 1 + ((-3/2) (-2/3))/(1!) + ((-3/2)(-3/2) - (1)/(2!) (-2/3)^2 + ... ` [`because ` From (2) ]
` rArr (1 - (2)/(3)) ^(-3/2) = 1 + 1 + (5 )/(2!3) + (5.7 ) /(3!3^2) + ... `
` rArr 3^(3//2) = 2 + (5)/(2!3) + (5.7 ) / (3 !3 ^ 2 ) + ... `
` 3 ^(3//2) = 2 + alpha `
` rArr (5 ) /2 ! 3 ) + (5.7 ) /( 3 ! 3 ^ 2 ) + ... = 3 `
` therefore alpha = 3 ^(3//2 ) - 2 `
` alpha + 2 = 3 ^(3//2 ) `
` rArr ( alpha + 2 ) ^2 = 3 ^ 2 `
` rArr alpha ^ 2 + 4 alpha = 27 - 4 `
` rArr alpha ^ 2 + 4 alpha = 23 `