Question

Let $\sum _{r=m}^{n-1}\frac{1}{r(r+1)}=\frac{1}{11}$ (where $m,n\in N$), then

• A. $5n$ is divisible by 11
• B. m is divisible by 10
• C. $mn=1100$
• D. Ordered pair $(m,n)$ is unique

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $5n$ is divisible by 11
B m is divisible by 10
C $mn=1100$
D Ordered pair $(m,n)$ is unique

$\sum _{r=m}^{n-1}(\frac{1}{r}-\frac{1}{r+1})=\frac{1}{m}-\frac{1}{n}=11$
$\Rightarrow n=\frac{11m}{11-m}\Rightarrow n=\frac{11(m-11+11)}{11-m}$
$n=\frac{121}{11-m}-11$
Hence $11-m$ is a divisor of 121, so $11-m=1$
$\Rightarrow m=10$
so $n=121-11=110$
Hence A, B, C, D