Let $\left(a+b x+c x^{2}\right)^{10}=\sum_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}$. If $p_{1}

Question

Let $\left(a+b x+c x^{2}ight)^{10}=\sum_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}$. If $p_{1}=20$ and $p_{2}=210$, then $2(a+b+c)$ is equal to\

Professor PC Thomas Classes · Answer

(c) : We have, $\left(a+b x+c x^{2}ight)^{10}$ $=\sum_{i=0}^{20} p_{i} x_{i}, a, b, c \in \mathbb{N}, p_{1}=20$ and $p_{2}=210$\ Now $\left(a+b x+c x^{2}ight)^{10}=p_{0}+p_{1} x+p_{2} x^{2}+\ldots \ldots$\ ${ }^{10} C_{1} a^{9} \cdot b=p_{1}$ and ${ }^{10} C_{2} a^{8} \cdot b^{2}+{ }^{10} C_{1} \cdot a^{9} \cdot c=p_{2}$\ $\Rightarrow \quad 10 b a^{9}=20$ and $45 b^{2} a^{8}+10 c \cdot a^{9}=210$\ $\Rightarrow b a^{9}=2$\ $\Rightarrow b=2$ and $a=1 \quad[\because a, b \in \mathbb{N}]$\ Now, $45 	imes(2)^{2}+10 \cdot c=210$\ $\Rightarrow \quad 10 c=210-180 \Rightarrow c=3$\ $	herefore \quad 2(a+b+c)=2(1+2+3)=2 	imes 6=12$\