Question

find the middle term of the expansion of [3-x³/6] whole raise to 8

Answer 1

$\fbox{ \mathtt{ \huge{ \fbox{SOLUTION :}}}}$

Here , the n = 8 (even)

So , the middle term of the given expansion is {(n+2)/2}th term i.e 5 term

We know that , the general term of an expansion (a + b)^n is given by

$\large \mathtt{ \underline{ \fbox{T_{r + 1} = {n}^{} C_{r} \times {(a)}^{n - r} \times {(b)}^{r} }}}$

For the 5th term , we have r + 1 = 5 , i.e r = 4

Therefore ,

$\mapsto \sf T_{4+ 1} = {8}^{} C_{4} \times {(3)}^{8- 4} \times {( \frac{ {( - x)}^{3} }{6} )}^{4} \\ \\ \mapsto \sf T_{4+ 1}= \frac{8!}{4!(8 - 4)!} \times {(3)}^{4} \times \frac{ {( - x)}^{12} }{ {(6)}^{4} } \\ \\ \mapsto \sf T_{4+ 1} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times 81 \times \frac{ {(x)}^{12} }{1296} \\ \\ \mapsto \sf T_{4+ 1} = 2 \times 7 \times 5 \times 81 \times \frac{ {(x)}^{12} }{1296} \\ \\ \mapsto \sf T_{4+ 1} = \frac{70 \times 81}{1296} {(x)}^{12} \\ \\ \mapsto \sf T_{4+ 1} = \frac{5670}{1296} {(x)}^{12}$

Hence , the middle term of given expansion is (5670/1296)(x)¹²