If the 2nd, 3rd and 4th terms in the expansion of `(x+a)^n` are 240, 720 and 1080 respectively, find `x \ and\ ndot`

Answer

B

Explanation

2nd term is `.^(n)C_(1)x^(n-1) a = 240" "(1)`
3rd term is `.^(n)C_(2)x^(n-2)a^(2) = 720" "(2)`
4th term is `.^(n)C_(3)x^(n-3)a^(3) = 1080 " "(3)`
Multiplying (1) and (3) and dividing by the square of (2), we get
`(.^(n)C_(1) xx .^(n)C_(3))/((.^(n)C_(2))^(3)) = (240 xx 1080)/((720)^(2))`
or `(nxxn(n-1)(n-2)(2!)^(2))/(n^(2)(n-1)^(2)xx3!) = 1/2`
or `4(n-2) = 3(n-1)" "( :' n ne 1)`
or `n = 5`
Putting `n = 5`, from (1) and (2), we get
`5x^(4)a = 240` and `10x^(3)a^(2) = 720`
`rArr ((5x^(4)a)^(2))/(10x^(3)a^(2)) = ((240)^(2))/(720)`
or `x^(5) = 32`
or `x = 32`
`:. a = (240)/(5x^(4)) = (48)/(2^(4)) = 3`
Hence, `x = 2, a = 3` and `n = 5`.
`(x-a)^(n) = (2-3)^(5) = -1`
Also, `(2+3)^(5) = 2^(5) + .^(5)C_(1)2^(4) xx 3+.^(5)C_(2)2^(3)xx3^(2) + .^(5)C_(3)2^(2) xx3^(3)+.^(5)C_(4)2xx3^(4)+.^(5)C_(5)3^(5)`
`= 32+240+720+1080+810+243`
Hence, least value of the term is 32.
Sum of odd numbered term is `32+720+810=1562`.