Question

Find the $7$th term of ${(3x^2-\frac{1}{3})}^{10}$

Answer 1

Using Binomial theorem

${(3x^2-\frac{1}{3})}^{10}=\sum _{n=0}^{10}$ ${}^{10}{C}_n(3x^2{)}^{10-n}{(-\frac{1}{3})}^n$

$T_{r+1}={}^{10}{C}_r(3x^2{)}^{10-r}{(-\frac{1}{3})}^r$

$7^{th}$ term of ${(3x^2-\frac{1}{3})}^{10}={}^{10}{C}_{7-1}(3x^2{)}^{10-6}{(-\frac{1}{3})}^6$

$={}^{10}{C}_6(3x^2{)}^4\frac{1}{3^6}$

$=\frac{10!}{6!4!}{3}^4\cdot x^8\frac{1}{3^6}$ [Using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$=\frac{10\times 9\times 8\times 7\times 6!}{6!4!}\frac{x^8}{3^2}$

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

$=\frac{70}{3}{x}^8$.