Question

Let $1+\displaystyle\sum_{ r =1}^{10}\left(3^{ r } \cdot{ }^{10} C _{ r }+ r \cdot{ }^{10} C _{ r }\right)=2^{10}\left(\alpha \cdot 4^5+\beta\right)$ where $\alpha, \beta \in N$ and $f ( x )= x ^2-2 x - k ^2+1$. If $\alpha, \beta$ lies between the roots of $f ( x )=0$, then find the smallest positive integral value of $k$.

Answer 1

We have $1+\sum_{ r =1}^{10}\left(3^{ r } \cdot{ }^{10} C _{ r }+ r \cdot{ }^{10} C _{ r }\right)=1+\sum_{ r =1}^{10} 3^{ r \cdot 10} C _{ r }+10 \sum_{ r =1}^{10}{ }^9 C _{ r -1}=1+4^{10}-1+10 \cdot 2^9$
$=4^{10}+5.2^{10}=2^{10}\left(4^5+5\right)=2^{10}\left(\alpha \cdot 4^5+\beta\right) \text {, so } \alpha=1 \text { and } \beta=5 $
$\text { Now } f(1)<0 \text { and } f(5)<0$
So $ f (1)<0 \Rightarrow - k ^2<0 \Rightarrow k \neq 0$
and $f (5)<0 \Rightarrow 16- k ^2<0 \Rightarrow k ^2-16>0 \Rightarrow k \in(-\infty, 4) \cup(4, \infty)$
Hence smallest positive integral value of $k = 5$