Let us consider the binomial expansion ` (1 + x)^(n) = sum_(r=0)^(n) a_(r) x^(r)`
where ` a_(4) , a_(5) "and " a_(6) ` are in AP , ( n ` lt ` 10 ). Consider another
binomial expansion of ` A = root (3)(2) + (root(4) (3))^(13n)` , the expansion of A
contains some rational terms ` T_(a1),T_(a2),T_(a3),...,T_(am)`
` (a_(1) lt a_(2) lt a_(3) lt ...lt a_(m))`
The value of ` a_(m)` is

Options

A. 87
B. 88
C. 89
D. 90

Answer

c

Explanation

`because a_(4) , a_(5) , a_(6) " i.e. """^(n)C_(4) , ""^(n)C_(5),""^(n)C_(6) ` are in AP , then
` 2. ""^(n)C_(5) = ""^(n)C_(4) + ""^(n)C_(6)`
`rArr 2= (""^(n)C_(4))/(""^(n)C_(5)) + (""^(n)C_(6))/(""^(n)C_(5)) = (5)/(n-5 + 1) + (n-6 +1)/(6)`
` rArr 2= (5)/(n-4) + (n-5)/(6)`
` rArr 12 n - 48 = 30 + n^(2) - 9n + 20 `
` rArr n^(2) - 21 n + 98 = 0 rArr n = 7,14 `
Hence , ` n = 7 " " [ because n lt 10]`
Also , `A = (root(3)(2) + root(4)(3))^(13n) = (2^(1//3) + 3^(1//4))^(91) `
` therefore T_(r+1) = ""^(91)C_(r) (2^(1//3)) ^(91-r) . (3^(1//4))^(r)`
` = ""^(91)C_(r) . 2^(91-r)/(3) . 3^(r//4)` ...(i)
From Eq . (i) , we get
` 0 le r le 91`
For rational terms , ` r = 4,16,28,40,52,64,76,88`
Rational terms are ` T_(5), T_(17), T_(29), T_(41), T_(53), T_(65), T_(77), T_(89) `
` therefore a_(m) = 89 `