Let a sequence an be defined by an=1n+1+1n+2+1n+3+…+13n, then which of the following is true statement

∵ an=1n+1+1n+2+1n+3+…+13nan=1n+1+1n+2+1n+3+…+1n+2nan=∑α=1m 1n+αa2=∑α=14 12+α=13+14+15+16=20+15+12+1060=5760=1920Now, an+1−an=1n+2+1n+3+…+13n+3 −1n+1+1n+2+…+13n=13n+1+13n+2+13n+3−1n+1=13n+1+13n+2−23(n+1)=9n2+15n+6+9n2+12n+3−18n2−18n−4(3n+1)(3n+2)(3n+3)=9n+5(3n+1)(3n+2)(3n+3)

Let a sequence an be defined by an=1n+1+1n+2+1n+3+…+13n, then which of the following is true statement

∵ an=1n+1+1n+2+1n+3+…+13nan=1n+1+1n+2+1n+3+…+1n+2nan=∑α=1m 1n+αa2=∑α=14 12+α=13+14+15+16=20+15+12+1060=5760=1920Now, an+1−an=1n+2+1n+3+…+13n+3 −1n+1+1n+2+…+13n=13n+1+13n+2+13n+3−1n+1=13n+1+13n+2−23(n+1)=9n2+15n+6+9n2+12n+3−18n2−18n−4(3n+1)(3n+2)(3n+3)=9n+5(3n+1)(3n+2)(3n+3)

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Let a sequence $\{a_{n}\}$ be defined by
$a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{3 n}$ , then which of the following is true statement

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$a_{2} = \frac{11}{12}$

$a_{2} = \frac{19}{20}$

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

$a_{n + 1} - a_{n} = \frac{- 2}{3 (n + 1)}$

answer is B.

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Detailed Solution

$∵ a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{3 n}$

$\begin{array}{l} a_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{n + 2 n} \\ a_{n} = \sum_{α = 1}^{m} \frac{1}{n + α} \end{array}$

$\begin{array}{l} a_{2} = \sum_{α = 1}^{4} \frac{1}{2 + α} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{20 + 15 + 12 + 10}{60} \\ = \frac{57}{60} = \frac{19}{20} \end{array}$

Now, $a_{n + 1} - a_{n} = (\frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{3 n + 3})$

$- (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{3 n})$

$\begin{array}{l} = \frac{1}{3 n + 1} + \frac{1}{3 n + 2} + \frac{1}{3 n + 3} - \frac{1}{n + 1} \\ = \frac{1}{3 n + 1} + \frac{1}{3 n + 2} - \frac{2}{3 (n + 1)} \\ = \frac{9 n^{2} + 15 n + 6 + 9 n^{2} + 12 n + 3 - 18 n^{2} - 18 n - 4}{(3 n + 1) (3 n + 2) (3 n + 3)} \\ = \frac{9 n + 5}{(3 n + 1) (3 n + 2) (3 n + 3)} \end{array}$