Question

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of `(root(4)2+1/(root(4)3))^n` is `sqrt(6):1```

Answer 1

`r+1^(th)` term in the given expression can be given as,
`T_(r+1) = C(n,r)(root4 2)^(n-r)(1/(root4 3))^r`
`:. T_5 = C(n,4)(root4 2)^(n-4)(1/(root4 3))^4`
Fifth term from end will be `T_(n-3)`
`:. T_(n-3) = C(n,n-4)(root4 2)^(4)(1/(root4 3))^(n-4)`
Now, we are given,
`T_5/T_(n-3) = sqrt6/1`
`:. (C(n,4)(root4 2)^(n-4)(1/(root4 3))^4)/(C(n,n-4)(root4 2)^(4)(1/(root4 3))^(n-4)) = sqrt6/1`
We know, `C(n,4) = C(n,(n-4))`
`:. (2^(n/4-1)3^(-1))/(2^1 3^(1-n/4)) = sqrt6/1`
`=>2^(n/4-2) 3^(n/4-2) = 2^(1/2)3^(1/2)`
`:. n/4-2 = 1/2`
`=>n/4 = 5/2=> n = 10`