Find the middle terms in the expansion of

    Q8.    \left(\frac{x}{3} + 9y \right )^{10}

Answers (1)
P Pankaj Sanodiya

As we know that the middle term in the expansion of  (a+b)^n when n is even is,

\left ( \frac{n}{2}+1 \right )^{th}\:term\:,

Hence the middle term of the expansion     \left(\frac{x}{3} + 9y \right )^{10}   is,

\left ( \frac{10}{2}+1 \right )^{th}\:term\:

Which is 6^{th}\:term

Now, 

As we know that the general  (r+1)^{th} term  T_{r+1} in the binomial expansion of  (a+b)^n  is given by 

T_{r+1}=^nC_ra^{n-r}b^r

So the 6^{th} term of the expansion of   \left(\frac{x}{3} + 9y \right )^{10} is

\\\Rightarrow T_6= T_{5+1}\\=^{10}C_5\left ( \frac{x}{3} \right )^{10-5}\left ( 9y \right )^5\\

=\left ( \frac{1}{3} \right )^5\times9^5\times^{10}C_5\times x^5y^5

=\left ( \frac{1}{3} \right )^5\times9^5\times\left ( \frac{10!}{5!5!} \right )\times x^5y^5

=\left ( \frac{1}{3^5} \right )\times9^5\times\left ( \frac{10\times9\times8\times7\times6}{5\times4\times3\times2} \right )\times x^5y^5

=61236x^5y^5

Hence the middle term of the expansion of  \left(\frac{x}{3} + 9y \right )^{10} is  61236x^5y^5.

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