Question

Find numerically the greatest term in the expansion of (2 + 3x)⁹, where x = `3/2`.

Answer 1

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 T_r+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`