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Question
In the following expansion, find the indicated term.
`(1/3 + "a"^2)^12`, 9th term
Solution
Here, a = `1/3`, b = a2, n = 12
For 9th term, r = 8
We have, tr+1 = nCr an–r .br
∴ t9 = `""^12"C"_8(1/3)^(12-8) ("a"^2)^8 = (12!)/(8!4!).(1/3)^4."a"^16`
= `(12 xx 11 xx 10 xx 9)/(4 xx 3 xx 2 xx 1) xx 1/81 xx "a"^16`
= `55/9"a"^16`
∴ 9th term in the expansion of `(1/3 + "a"^2)^12` is `55/9"a"^16`
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