If the sixth term from the beginning is equal to the sixth term from the end in the expansion of (21/5+3−1/5)n, then the sum of the rational terms in the expansion is
A
13499
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B
15499
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C
17599
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D
11199
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Solution
The correct option is B15499 In the expansion of (21/5+3−1/5)n, Tr+1=nCr(21/5)n−r(3−1/5)r
Given that T6=Tn−4 ⇒nC5(2n/56)=nCn−5(63n/5) ⇒6n/5=36 ⇒n=10
∴Tr+1=10Cr⋅(2)10−r5⋅(3)−r5
The term will be rational only when r=0,5,10 ∴ Sum of rational terms =T1+T6+T11 =10C0⋅22+10C5⋅23+10C10⋅19 =4+168+19 =15499