For any n∈ℕ, 1/2.5+1/5.8+…+1/((3n-1)(3n+2))=

Correct answer: (A) n/(6n+4). S_n=1/2.5+1/5.8+…+1/((3n-1)(3n+2)), n∈ℕ T_n=1/(3n-13n+2) ⇒T_n=1/31/(3n-1)-1/(3n+2) ∴T_1=1/31/2-1/5 T_2=1/31/5-1/8 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ T_n=1/31/(3n-1)-1/(3n+2) S_n=T_1+T_2+T_3+----+T_n ∴S_n=1/31/2-1/(3n+2)=n/(6n+4)

Q43 / 160 · TS EAMCET 2021 (04 Aug Shift 1) · Mathematics

For any $n \in ℕ, \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} =$

Correct answer
A
$\frac{n}{6 n + 4}$
B
$\frac{n^{2}}{6 n + 4}$
C
$\frac{1}{2} \cdot \frac{n^{2}}{6 n + 4}$
D
$\frac{n}{6 n^{2} + 4}$

Solution

$S_{n} = \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3 n - 1) (3 n + 2)}, n \in ℕ$

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

$\Rightarrow T_{n} = \frac{1}{3} [\frac{1}{3 n - 1} - \frac{1}{3 n + 2}]$

$∴ T_{1} = \frac{1}{3} [\frac{1}{2} - \frac{1}{5}]$

$T_{2} = \frac{1}{3} [\frac{1}{5} - \frac{1}{8}]$

$\begin{matrix} ⋮ \\ ⋮ \\ ⋮ \end{matrix} \begin{matrix} ⋮ \\ ⋮ \\ ⋮ \end{matrix}$

$T_{n} = \frac{1}{3} [\frac{1}{3 n - 1} - \frac{1}{3 n + 2}]$

$S_{n} = T_{1} + T_{2} + T_{3} + - - - - + T_{n}$

$∴ S_{n} = \frac{1}{3} [\frac{1}{2} - \frac{1}{3 n + 2}] = \frac{n}{6 n + 4}$

Hence, the correct answer is option: 'A'.

For any n∈ℕ, 12.5+15.8+…+1(3n-1)(3n+2)=

Solution

Relevant Questions

For any $n \in ℕ, \frac{1}{2.5} + \frac{1}{5.8} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} =$