# Find the middle term in the expansion of $$(3x – {x^3\over 6})^7$$.

## Solution :

The given expression is $$(3x – {x^3\over 6})^7$$.

Here n = 7, which is an odd number.

So, $$({7 + 1\over 2}$$) th and $$({7 + 1\over 2} + 1)$$ th i.e.  4th and 5th terms are two middle terms.

Now, $$T_{4}$$ = $$T_{3 + 1}$$ = $$^{7}C_{3}$$ $$(3x)^{7 – 3}$$ $$(-{x^3\over 6})^{3}$$

$$\implies$$ $$T_{4}$$ = $$(-1)^3$$ $$^{7}C_{3}$$ $$(3x)^{4}$$ $$({x^3\over 6})^{3}$$

$$\implies$$ $$T_{4}$$ = -35 $$\times$$ 81 $$x^4$$ $$\times$$ $$x^9\over 216$$ = -$$105x^{13}\over 8$$

and,  $$T_{5}$$ = $$T_{5 + 1}$$ = $$^{7}C_{4}$$ $$(3x)^{7 – 4}$$ $$(-{x^3\over 6})^{4}$$

$$\implies$$ $$T_{5}$$ = $$^{7}C_{4}$$ $$(3x)^{3}$$ $$({x^3\over 6})^{4}$$

$$\implies$$ $$T_{5}$$= 35 $$\times$$ 27 $$x^3$$ $$\times$$ $$x^{12}\over 1296$$ = $$35x^{15}\over 48$$

Hence, the middle terms are -$$105x^{13}\over 8$$ and $$35x^{15}\over 48$$.

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