Find the middle terms in the expansion `(3x-(x^3)/6)^7` .

Explanation

The general term in the given expansion is given by
`T_(r+1) = (-1) ^(r) xx .^(7) C _(r) xx (3x)^(7-r) xx ((x^(3))/6)^(r)`
`rArr T _(r+1) = (-1) ^(r) xx .^(7)C_(r) xx 3^((7-2r)) xx 2 ^(-r) xx x^(7+2r)" "` …(i)
Clearly, the given expansion has 8 terms.
So, it has two middle terms, namely `(8/2)` th , i. e. , 4th and 5th.
Now, `T _(4) = T _(3+1)`
`=(-1)^(3) xx .^(7) C _(3) xx 3 ^((7-6)) xx 2 ^(-3) xx x^(13)" "` [putting r = 3 in (i) ]
`= - (( 7 xx 6 xx 5 )/(3 xx 2 xx 1 ) xx3^(1) xx 1 / 2^(3) xx x ^(13))`
`=-((35 xx 3 xx x^(13))/8) = (-105x^(13))/8.`
And, `T_(5)= T_((4+1))`
`(-1)^(4) xx .^(7) C_(4) xx 3^((7-8)) xx 2 ^ (-4) xx x^(15)" "` [putting r = 4 in (i)]
`= .^(7) C _(3) xx 3^(-1) xx 2 ^(-4) xx x^(15)" "[ :' .^(7)C_(4) = .^(7) C _((7-4)) = .^(7) C _(3)]`
`((7 xx 6 xx 5 )/( 3xx 2 xx 1 ) xx 1/3 xx 1/(2^(4)) xx x^(15)). (35x^(15))/48.`
Hence, the required middle terms are `(-105 x^(13))/8 and (35x^(15))/48.`