# Find the 9th term in the expansion of $$({x\over a} – {3a\over x^2})^{12}$$.

## Solution :

We know that the (r + 1)th term or general term in the expansion of $$(x + a)^n$$ is given by

$$T_{r + 1}$$ = $$^{n}C_r x^{n – r} a^r$$

In the expansion of $$({x\over a} – {3a\over x^2})^{12}$$, we have

$$T_{9}$$ = $$T_{8 + 1}$$ = $$^{12}C_8 ({x\over a})^{12 – 8} ({-3a\over x^2})^8$$

$$\implies$$ $$T_{9}$$ = $$^{12}C_8 ({x\over a})^4 ({-3a\over x^2})^8$$

$$\implies$$ $$T_{9}$$ = $$^{12}C_4$$ $$3^8 a^4\over x^{12}$$ = $$(^{12}C_4 x^{-12} a^4)3^8$$

### Similar Questions

Find the middle term in the expansion of $$({2\over 3}x^2 – {3\over 2x})^{20}$$.

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

Find the 10th term in the binomial expansion of $$(2x^2 + {1\over x})^{12}$$.

Which is larger $$(1.01)^{1000000}$$ or 10,000?

By using binomial theorem, expand $$(1 + x + x^2)^3$$.