Question

Let a sequence $\left\{a_n\right\}$ be defined by $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+.....+\frac{1}{3n}$, then

• A. $a_2=\frac{7}{12}$
• B. $a_2=\frac{19}{20}$
• C. $a_{n+1}-a_n=\frac{9n+5}{(3n+1)(3n+2)(3n+3)}$
• D. $a_{n+1}-a_n=\frac{-2}{3(n+1)}$

Answer 1

Answer: (B), (C)

$a_2=\frac{19}{20}$
$a_{n+1}-a_n=\frac{9n+5}{(3n+1)(3n+2)(3n+3)}$

$a_n$ is sum of reciprocals of natural numbers starting from $n+1$ and ending at $3n$.
$\therefore a_2=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}$
$\therefore a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{3n}+\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}$
$\Rightarrow a_{n+1}-a_n=\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+3}-\frac{1}{n+1}$
$=\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}$
$=\frac{(3n+2)(3n+3)}{(3n+1)(3n+2)(3n+3)}+\frac{(3n+1)(3n+3)}{(3n+1)(3n+2)(3n+3)}-\frac{2(3n+1)(3n+2)}{(3n+1)(3n+2)(3n+3)}$
$=\frac{9n+5}{(3n+1)(3n+2)(3n+3)}$
Hence, options 'B' and 'C' are correct.