Prove the following 1-2-5-1-5-8-1-8-11-1-3 n-13 n-2-n-6 n-4

1/2.5+1/5.8+1/8.11+⋯ +1/(3n 1)(3n+2)=n/6n+4 So, 1/2.5=1/3 ( 1/5 1/8 ) 1/(3n 1)(3n+2)=1/3 ( 1/3n 1 1/3n+1 ) So, it is 1/3{1/2 1/3n+2} =1/3{3n+2 2/2(3n+2)}= ( 3n/ ...

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Standard XII

Mathematics

Definition of Vector

Prove the fol...

Question

Prove the following:
$\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}$

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Solution

$\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}$
So,
$\frac{1}{2.5} = \frac{1}{3} (\frac{1}{5} - \frac{1}{8})$
$\frac{1}{(3 n - 1) (3 n + 2)} = \frac{1}{3} (\frac{1}{3 n - 1} - \frac{1}{3 n + 1})$
So, it is $\frac{1}{3} {\frac{1}{2} - \frac{1}{3 n + 2}}$
$= \frac{1}{3} {\frac{3 n + 2 - 2}{2 (3 n + 2)}} = (\frac{3 n}{6 n + 4}) \times \frac{1}{3}$
$= \frac{n}{6 n + 4}$

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Similar questions

$\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}$

$\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . . + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}$

Prove the following by using the principle of mathematical induction for all n ∈ N:

Q. Prove the following by using the principle of mathematical induction for all

n \in N : \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . . . . + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{(6 n + 4)}

Q. If

\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . . . . . . . . . . . . + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{a n + 4} .

Find

a

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Prove the following: 12.5+15.8+18.11+⋯+1(3n−1)(3n+2)=n6n+4

12.5+15.8+18.11+⋯+1(3n−1)(3n+2)=n6n+4 So, 12.5=13(15−18) 1(3n−1)(3n+2)=13(13n−1−13n+1) So, it is 13{12−13n+2} =13{3n+2−22(3n+2)}=(3n6n+4)×13 =n6n+4

Prove the following:
$\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + \dots + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4}$