Find the middle terms in the expansion of $$( \dfrac{x}{3}+9y)^{10}$$

Asked by Abhisek | 1 year ago |  67

##### Solution :-

It is known that in the expansion of (a+b)n in n is even, then the middle term

$$( \dfrac{n}{2}+1)^{th}$$

Therefore, the middle term in the expansion of

$$( \dfrac{10}{2}+1)^{th}$$ = 6th term

$$T_4=T_{5+1}$$

$$^{10}C_5(\dfrac{x}{3})^{10-5}$$

$$(9y)^5=\dfrac{10!}{5!5!}\times \dfrac{x^5}{3^5}\times 9^5\times y^5$$

$$\dfrac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 5!}$$

$$\times \dfrac{1}{3^5}\times 3^{10}\times x^5\times y^5$$

$$252\times 3^5\times x^5\times y^5$$

$$=61236x^5y^5$$

Thus, the middle term in the expansion of

$$=61236x^5y^5$$

Answered by Pragya Singh | 1 year ago

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