Question

Find the middle term in the expansion of:

$\left(i\right){(3x-\frac{x^3}{6})}^9$

$\left(ii\right){(2x^2-\frac{1}{x})}^7$

$\left(iii\right){(3x-\frac{2}{x^2})}^{15}$

$\left(iv\right){(x^4-\frac{1}{x^3})}^{11}$

Answer 1

$\left(i\right){(3x-\frac{x^3}{6})}^9$

Here, n =0, which is odd number

$\therefore {\left(\frac{9+1}{2}\right)}^{th}and{(\frac{9+1}{2}+1)}^{th}i.e.,5^{th},6^{th}$ term are the middle term.

Here, the term formula is

$T_5=T_{4+1}=(-1{)}^4{}^9{C}_4(3x{)}^5{\left(\frac{x^3}{6}\right)}^4$

$={}^9{C}_4\frac{3^5}{6^4}\times x^5\times x^{12}$

$=\frac{9\times 8\times 7\times 6\times 3^5}{4\times 3\times 2\times 3^4\times 2^4}{x}^{17}$

$=\frac{189}{8}{x}^{17}$

$T_6=T_{5+1}=(-1{)}^5{}^9{C}_5(3x{)}^4{\left(\frac{x^3}{6}\right)}^5$

$=-\frac{9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times \frac{3^4}{6^5}\times x^4\times x^{15}$

$=-\frac{9\times 8\times 7\times 6\times 3^4}{5\times 4\times 3\times 2\times 3^5\times 2^5}{x}^{19}$

$=\frac{-21}{16}{x}^{19}$

$\left(ii\right){(2x^2-\frac{1}{x})}^7$

Here, n=7, which is odd

$\therefore {\left(\frac{7+1}{2}\right)}^{th}and{(\frac{7+1}{2}+1)}^{th}=4^{th},5^{th}$ term are middle term or ${(2x^2-\frac{1}{x})}^7$

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_4=T_{3+1}=(-1{)}^3{}^7{C}_3(2x^2{)}^{7-3}{\left(\frac{1}{x}\right)}^3$

$=-{}^7{C}_3\frac{2^4{x}^8}{x^3}$

$=-560x^5$

$T+5=T_{4+1}=(-1{)}^4{}^7{C}_4(2x^2{)}^{7-4}{\left(\frac{1}{x}\right)}^4$

$={}^7{C}_4\frac{2^3{x}^6}{x^4}$

$={}^7{C}_4\frac{7\times 6\times 5\times 8}{3\times 2}x62$

$=280x^2$

(iii) Given:

n, i,e. 15 is an odd number .

Thus, the middle terms are $\left(\frac{15+1}{2}\right)thand\left(\frac{15+1}{2}\right)th$ i.e. 8th and 9th.

Now,

$T_8=T_{7+1}$

$={}^{15}{C}_7(3x{)}^{15-7}{(-\frac{2}{x^2})}^7,$

$=-\frac{15\times 14\times 13\times 12\times 11\times 10\times 9}{7\times 6\times 5\times 4\times 3\times 2}\times 3^8\times 2^7{x}^{8-14}$

$=\frac{-6435\times 3^8\times 2^7}{x^6}$

And,

$T_9=T_{8+1}$

$={}^{15}{C}_8(3x{)}^{15-8}{\left(\frac{-2}{x^2}\right)}^8$

$=\frac{15\times 14\times 13\times 12\times 11\times 10\times 9}{7\times 6\times 5\times 4\times 3\times 2}\times 3^7\times 2^8{x}^{7-16}$

$=\frac{6435\times 3^7\times 2^8}{x^9}$

$\left(iv\right){(x^4-\frac{1}{x^3})}^{11}$

Here, n=11, which is odd number

$\therefore {\left(\frac{11+1}{2}\right)}^{th}$ and ${(\frac{11+1}{2}+1)}^{th}=6^{th},7^{th}$ term are the middle terms in ${(x^4-\frac{1}{x^3})}^{11}$

The term formula is

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_6=T_{5+1}=(-1{)}^5{}^{11}{C}_5(x^4{)}^{11-5}{\left(\frac{1}{x^3}\right)}^5$

$=-{}^{11}{C}_5{x}^{24}\frac{1}{x^{15}}$

$=\frac{-11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2\times 1}{x}^9$

$=-11\times 3\times 2\times 7x^9$

$=-462x^9$

$T_7=T_{6+1}=(-1{)}^6{}^{11}{C}_6(x^4{)}^{11-6}{\left(\frac{1}{x^3}\right)}^6=462\frac{x^{20}}{x^{18}}$

$=462x^2$