Question

find the middle term in The expansion of (x-1/2y)^10​

Answer 1

Answer:

-63/8 x⁵y⁵

Step-by-step explanation:

By the binomial expansion, the middle term is

$\displaystyle\binom{10}{5}x^5\bigl(-\tfrac{y}{2}\bigr)^5\\\\=\frac{-1}{2^5}\binom{10}{5}x^5y^5\\\\=\frac{-1\times10\times9\times8\times7\times6}{2^5\times5\times4\times3\times2\times1}x^5y^5\\\\=\frac{-9\times7}{2^3}x^5y^5\\\\=-\tfrac{63}{8}x^5y^5$

Answer 2

The middle term in the expansion of $(x-\dfrac{1}{2}y)^{10}$ is $\dfrac{63}{8}x^5y^5$ .

Explanation:

Given expression : $(x-\dfrac{1}{2}y)^{10}$

Compare to $(a+b)^n$ , we have $a=x ,\ b=\dfrac{1}{2}y,\ \&\ n=10$

Since n= 10 , the total number of terms = n+1 =10+1=11 , which is odd.

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

$(r+1)^{th}$ term = $T(r+1)=^{n}C_ra^{n-r}b^r$

Put values of n=10 , r=5, a and b, we get

$T(5+1)=^{10}C_5x^{10-5}(\dfrac{1}{2}y)^5\\\\ T(6)=\dfrac{10!}{5!5!}x^5(\dfrac{1}{2})^5y^5\\\\=\dfrac{63}{8}x^5y^5$

Hence, the middle term in the expansion of $(x-\dfrac{1}{2}y)^{10}$ is $\dfrac{63}{8}x^5y^5$ .

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