Question

An equation $a_0+a_{1}x+a_2{x}^2+...+a_{99}{x}^{99}+x^{100}=0$ has root ${}^{99}{C}_0,{}^{99}{C}_1,{}^{99}{C}_2,...{}^{99}{C}_{99}$ then

• A. $({}^{99}{C}_0{)}^2+({}^{99}{C}_1{)}^2+({}^{99}{C}_2{)}^2+...({}^{99}{C}_{99}{)}^2=(a_{99}{)}^2-2a_{98}$
• B. $({}^{99}{C}_0{)}^2+({}^{99}{C}_1{)}^2+({}^{99}{C}_2{)}^2+...({}^{99}{C}_{99}{)}^2=(a_{99}{)}^2+2a_{98}$
• C. $a_{98}=2^{197}+\frac{1}{2}{}^{198}{C}_{99}$
• D. $a_{98}=2^{197}-\frac{1}{2}{}^{198}{C}_{99}$

Answer 1

Answer: (A), (D)

The correct options are
A $({}^{99}{C}_0{)}^2+({}^{99}{C}_1{)}^2+({}^{99}{C}_2{)}^2+...({}^{99}{C}_{99}{)}^2=(a_{99}{)}^2-2a_{98}$
D $a_{98}=2^{197}-\frac{1}{2}{}^{198}{C}_{99}$

sum of the roots ${}^{99}{C}_0+{}^{99}{C}_1+...+{}^{99}{C}_{99}=-a_{99}{a}_{99}=-2^{99}...\left(1\right)$
sum of the product of the two roots at a time is
$a_{98}=\sum _{0\le i<}\sum _{j\le 99}{}^{99}{C}_i{}^{99}{C}_j=\frac{1}{2}\sum _{i=0}^{99}\sum _{j=0}^{99}{}^{99}{C}_i{}^{99}{C}_j-\frac{1}{2}\sum _{i=0}^{99}{\left({}^{99}{C}_i\right)}^2=2^{197}-\frac{1}{2}{}^{198}{C}_{99}...\left(2\right)(\because (1+x{)}^{2n}=(1+x{)}^n(x+1{)}^n\text{comparing the coefficients of}{x}^n)$
From $\left(1\right)$ and $\left(2\right)$
$({}^{99}{C}_0{)}^2+({}^{99}{C}_1{)}^2+({}^{99}{C}_2{)}^2+...({}^{99}{C}_{99}{)}^2=(a_{99}{)}^2-2a_{98}$