Question

If $(1+x+x^2{)}^{100}=\sum _{r=0}^{200}{a}_r{x}^r$, then which of the following is/are true?

• A. $a_0+a_1+.......+a_{99}=\frac{3^{100}-a_{100}}{2}$
• B. $a_0+a_1+.......+a_{99}=\frac{3^{101}-a_{99}}{2}$
• C. $a_0+a_1+.......+a_{100}=\frac{3^{100}+a_{100}}{2}$
• D. $a_0+a_1+.......+a_{100}=\frac{3^{100}-a_{100}}{2}$

Answer 1

Answer: (A), (C)

The correct options are
A $a_0+a_1+.......+a_{99}=\frac{3^{100}-a_{100}}{2}$
C $a_0+a_1+.......+a_{100}=\frac{3^{100}+a_{100}}{2}$

$(1+x+x^2{)}^{100}=\sum _{r=0}^{200}{a}_r{x}^r....(1)$
$\sum _{r=0}^{200}{a}_r{\left(\frac{1}{x}\right)}^r=(1+\frac{1}{x}+\frac{1}{x^2}{)}^{100}=\frac{(1+x+x^2{)}^{100}}{x^{200}}\sum _{r=0}^{200}{a}_r{x}^{200-r}=(1+x+x^2{)}^{100}$

Replacing $r$ by $200-r$ in the equation $\left(1\right)$
$(1+x+x^2{)}^{100}=\sum _{r=0}^{200}{a}_{200-r}{x}^{200-r}....(3)$

From the equation $\left(2\right)$ and $\left(3\right)$
$a_r=a_{200-r}{a}_0+a_1+.....+a_{200}=3^{100}$
So,$2(a_0+a_1+.....+a_{99})+a_{100}=3^{100}{a}_0+a_1+.....+a_{99}=\frac{3^{100}-a_{100}}{2}$
similarly if we add $a_{100}$ both side we get
$a_0+a_1+.......+a_{100}=\frac{3^{100}+a_{100}}{2}$