Question

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

Answer 1

Answer: 2.01

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