Question

Find the middle term of the following (2x-3/x)*20

Answer 1

Answer:

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Answer 2

Answer: $^{20}C_{10}\cdot2^{10}\cdot3^{10}$

Step-by-step explanation:

When we compare $(2x-\dfrac{3}{x})^{20}$ to $(a+b)^n$ , we get n=20 which is even .

Middle term in any binomial expansion = $(\dfrac{n}{2}+1)^{th}\text{term}$

i.e. Middle term  = $(\dfrac{20}{2}+1)^{th}\text{term}=11^{th}\text{term}$

∵ (r+1) th term of binomial = $T_{r+1}=^nC_ra^{n-r}b^r$

Then, 11th term of  $(2x-\dfrac{3}{x})^{20}$  will be :-

$T_{11}=T_{10+1}=^{20}C_{10}(2x)^{20-10}(-\dfrac{3}{x})^{10}\\\\=^{20}C_{10}(2^{10}x^{10})\times(\dfrac{3^{10}}{x^{10}})\\\\=^{20}C_{10}\cdot2^{10}\cdot3^{10}$

Hence, middle term of $(2x-\dfrac{3}{x})^{20}$$=^{20}C_{10}\cdot2^{10}\cdot3^{10}$