Question

The number of distinct terms in the expansion of ${(x^2+x+1+\frac{1}{x})}^5$ is N, then the value of $N^2$ is

• A. $225$
• B. $0$
• C. $256$
• D. $400$

Answer 1

Answer: (C)

$256$

${(x^2+x+1+\frac{1}{x})}^5$

$={\left(x\right(x+1)+\frac{1}{x}(x+1\left)\right)}^5$

$={((x+\frac{1}{x})(x+1))}^5$

$={(x+\frac{1}{x})}^5.{(x+1)}^5$

$=[{x^5+{}^{5}C}_1.x^4.\frac{1}{x}+{{}^{5}C}_2.x^3.\frac{1}{x^2}+{{}^{5}C}_3.x^2.\frac{1}{x^3}+{{}^{5}C}_4.x^1.\frac{1}{x^4}+\frac{1}{x^5}]\times$

$[x^5+{{}^{5}C}_1.x^4+{{}^{5}C}_2.x^3+{{}^{5}C}_3.x^2+{{}^{5}C}_4.x^1+1]$

$=[{x^5+{}^{5}C}_1.x^3+{{}^{5}C}_2.x+{{}^{5}C}_3.1/x+{{}^{5}C}_4.1{/x}^3+1/x^5]\times [{x^5+{}^{5}C}_1.x^4+{{}^{5}C}_2.x^3+{{}^{5}C}_3.x^2+{{}^{5}C}_4.x^1+1]$

$\therefore$ Different terms of $x$ include ,

$=x^{10},x^9,x^8,x^7,x^6,x^5,x^4,x^3,x^2,x^1,x^0,x^{-1},x^{-2},x^{-3},x^{-4}and{x}^{-5}$

Total 16 items.

$\Rightarrow N=16$

$\Rightarrow N^2=256$.

Hence the answer is $256.$