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Let us consider the binomial expansion ...

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 ` sum_(i=1)^(n) a_(i)` is

A

63

B

127

C

255

D

511

Text Solution

Verified by Experts

The correct Answer is:
b
`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)
`sum_(i=1)^(n)a_(i) = sum_(i=1)^(7) a_(i) = a_(1) + a_(2) + a_(3) + ...+ a_(7) `
` = ""^(7)C_(1) +""^(7)C_(2) + ""^(7)C_(3) + ...+ ""^(7)C_(7) = 2^(7) - 1 = 127`
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