(1 /2.5) + (1 /5.8)+ (1 /8.11)+.........(1/(3n-1)(3n+2)) =

Question

(1 /2.5) + (1 /5.8)+ (1 /8.11)+.........(1/(3n-1)(3n+2)) = (A)  (n/6n+3) (B)  (n/6n-4) (C)  (n+1/6n+4) (D)  (n/6n+4)

Tardigrade Admin · Accepted Answer

Given , (1/2.5 ) + (1/5.8) + (1/8.11) +... + (1/(3n-1)(3n+2)) = (1/3) [((1/2) - (1/5)) + ((1/5) -(1/8)) + ((1/8) - (1/11))+ .... + ((1/3n-1) - (1/3n+2))] = (1/3) [(1/2) - (1/3n+2)] = (3n+2-2/6(3n+2)) = (n/6n+4)

Q. $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ......... \frac{1}{( 3 n - 1 ) ( 3 n + 2 )}$ =

$\frac{n}{6 n + 3}$

$\frac{n}{6 n - 4}$

$\frac{n + 1}{6 n + 4}$

$\frac{n}{6 n + 4}$

Q. 2.51​+5.81​+8.111​+.........(3n−1)(3n+2)1​=

6n+3n​

6n−4n​

6n+4n+1​

6n+4n​

Given , 2.51​+5.81​+8.111​+...+(3n−1)(3n+2)1​ =31​[(21​−51​)+(51​−81​)+(81​−111​)+....+(3n−11​−3n+21​)] =31​[21​−3n+21​] =6(3n+2)3n+2−2​=6n+4n​

Q. $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ......... \frac{1}{( 3 n - 1 ) ( 3 n + 2 )}$ =

$\frac{n}{6 n + 3}$

$\frac{n}{6 n - 4}$

$\frac{n + 1}{6 n + 4}$

$\frac{n}{6 n + 4}$