Question

The number of terms in the expansion of ${\left(\frac{x^3+1+x^6}{x^3}\right)}^{{\sum }_n}$, (where n $ϵ$ N) is

• A. $n^2+n+1$
• B. $\sum n+1$
• C. ${}^{{\sum }_n+2}{C}_2$
• D. $2n+1$

Answer 1

The correct option is A $n^2+n+1$

Let $\sum n=\lambda$
${\left(\frac{x^3+1+x^6}{x^3}\right)}^{{\sum }_n}={(1+(x^3+\frac{1}{x^3}))}^{\lambda }$
$=1+{}^{\lambda }{C}_1(x^3+\frac{1}{x^3})+{}^{\lambda }{C}_2{(x^3+\frac{1}{x^3})}^2+\cdots +{}^{\lambda }{C}_{\lambda }{(x^3+\frac{1}{x^3})}^{\lambda }$
On expanding each term, two dissimilar terms are added in the expansion.
For eg. ${(x^3+\frac{1}{x^3})}^2=x^6+\frac{1}{x^6}+2$
${(x^3+\frac{1}{x^3})}^3=x^9+\frac{1}{x^9}+3(x^3+\frac{1}{x^3})$
Hence total number of terms
$=1+\underset{\lambda times}{2+2+\cdots +2}$
$=1+2\lambda$
$=1+2\sum n=1+\frac{2n(n+1)}{2}$
$=1+n+n^2$