Question

If $\displaystyle \sum _{k = 0}^{100}\left(\frac{k}{k + 1}\right)\left(\_{}^{100}\text{C}_{k}^{}\right)=\frac{a \cdot 2^{100} + b}{c}$ , where $a,b,c\in N$ , then the least value of $\left(\frac{a + b + c}{100}\right)$ is equal to

Answer 1

Let $S=\displaystyle \sum _{k = 0}^{100} \left(\frac{k}{k + 1}\right) \left(\left(\_{}^{100}C\right)_{k}\right)=\displaystyle \sum _{k = 0}^{100} \left(\frac{k + 1 - 1}{k + 1}\right) \left(\_{}^{100}C\right)_{k}$
$=\displaystyle \sum _{k = 0}^{100} \_{}^{100}C_{k}-\displaystyle \sum _{k = 0}^{100} \frac{1}{k + 1}\_{}^{100}C_{k}$
$=2^{100}-\frac{1}{101}\displaystyle \sum _{k = 0}^{100} \frac{101}{k + 1}\_{}^{100}C_{k}$
$=2^{100}-\frac{1}{101}\displaystyle \sum _{k = 0}^{100} \_{}^{101}C_{k + 1}$
$=2^{100}-\frac{1}{101}\left(2^{101} - 1\right)$
$= \frac{101 \cdot 2^{100} - 2^{101} + 1}{101} = \frac{99 \cdot 2^{100} + 1}{101}$
So, $a=99,b=1,c=101$
Hence, $\left(\frac{a + b + c}{100}\right)_{least}=2.01$