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Question
Find numerically the greatest term in the expansion of (2 + 3x)9, where x = `3/2`.
Solution
We have (2 + 3x)9 = `2^9 (1 + (3x)/2)^9`
Now, `("T"_(r + 1))/"T"_r = (2^9 [""^9"C"_r ((3x)/2)^r])/(2^9 [""^9"C"_(r - 1) ((3x)/2)^(r - 1)]`
= `(""^9"C"_r)/(""^9"C"_(r - 1)) |(3x)/2|`
= `9/(r(9 - r)) * ((r - 1)(10 - r))/9 |(3x)/2|`
= `(10 - r)/r |(3x)/2|`
= `(10 - r)/r (9/4)`
Since x = `3/2`
Therefore, `("T"_(r + 1))/"T"_r ≥ 1`
⇒ `(90 - 9r)/(4r) ≥ 1`
⇒ 90 – 9r ≥ 4r ....(Why)
⇒ r ≤ `90/13`
⇒ r ≤ 6 `12/13`
Thus the maximum value of r is 6
Therefore, the greatest term is Tr+1 = T7
Hence, T7 = `2^9 [""^9"C"_6 ((3x)/2)^6]`
Where x = `3/2`
= `2^9 * ""^9"C"_6 (9/4)^6`
= `2^9 * (9 xx 8 xx 7)/(3 xx 2 xx 1) (3^12/2^12)`
= `(7 xx 3^13)/2`
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