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  • Find the middle terms in the expansion of ( 3 - x ^ 3 / 6) ^7

Find the middle terms in the expansion of

    Q7.    \left(3 - \frac{x^3}{6} \right )^7

Answers (1)

best_answer

As we know that the middle  terms in the expansion of  (a+b)^n when n is odd are,

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

Hence the middle term of the expansion    \left(3 - \frac{x^3}{6} \right )^7   are 

\left ( \frac{7+1}{2} \right )^{th}\:term\: \:and\:\:\left ( \frac{7+1}{2}+1 \right )^{th}\:term

Which are 4^{th}\:term\:and\:\:5^{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 4^{th} term of the expansion of   \left(3 - \frac{x^3}{6} \right )^7 is

\\\Rightarrow T_4= T_{3+1}\\=^{7}C_3(3)^{7-3}\left (- \frac{x^3}{6} \right )^3\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times^{7}C_3\times x^{9}\\=\left ( -\frac{1}{6} \right )^3\times 3^4\times\frac{7!}{3!4!}\times x^{9} \\=-\frac{3\times3\times3\times3}{6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^9

=-\frac{105}{8}x^9

And the 5^{th} Term of the expansion of   \left(3 - \frac{x^3}{6} \right )^7  is,

\\\Rightarrow T_5= T_{4+1}\\=^{7}C_4(3)^{7-4}\left (- \frac{x^3}{6} \right )^4\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times^{7}C_4\times x^{12}\\=\left ( -\frac{1}{6} \right )^4\times 3^3\times\frac{7!}{3!4!}\times x^{12} \\=\frac{3\times3\times3}{6\times6\times6\times6}\times\frac{7\times 6\times5}{3\times2}\times x^{12}

=\frac{35}{48}x^{12}

Hence the middle terms of the expansion of given expression are

-\frac{105}{8}x^9\:and\:\frac{35}{48}x^{12}. 

 

Posted by

Pankaj Sanodiya

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