12-5-15-8-18-11-13 n-13 n-2-n 6 n-4

Let P(n) be the given statement. Now, P(n) #160;=12.5+15.8+18.11+...+1(3n 1)(3n+2)=n6n+4Step #160;1:P(1) #160;=12.5=110=16+4Hence, #160;P(1) #160;is ...

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Standard XII

Mathematics

Statement

12.5+15.8+18....

Question

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

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Solution

Let P(n) be the given statement.
Now,
$P (n) = \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . + \frac{1}{(3 n - 1) (3 n + 2)} = \frac{n}{6 n + 4} Step 1 : P (1) = \frac{1}{2.5} = \frac{1}{10} = \frac{1}{6 + 4} Hence, P (1) is true . Step 2 : Let P (m) be true . Then, \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . + \frac{1}{(3 m - 1) (3 m + 2)} = \frac{m}{6 m + 4} T o prove : P (m + 1) is true . i . e ., \frac{1}{2.5} + \frac{1}{5.8} + . . . + \frac{1}{(3 m + 2) (3 m + 5)} = \frac{m + 1}{6 m + 10} T h u s, w e h a v e : \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + . . . + \frac{1}{(3 m - 1) (3 m + 2)} = \frac{m}{6 m + 4} \Rightarrow \frac{1}{2.5} + \frac{1}{5.8} + . . . + \frac{1}{(3 m - 1) (3 m + 2)} + \frac{1}{(3 m + 2) (3 m + 5)} = \frac{m}{6 m + 4} + \frac{1}{(3 m + 2) (3 m + 5)} [Adding \frac{1}{(3 m + 2) (3 m + 5)} to both sides] \Rightarrow \frac{1}{2.5} + \frac{1}{5.8} + . . . + \frac{1}{(3 m + 2) (3 m + 5)} = \frac{3 m^{2} + 5 m + 2}{2 (3 m + 2) (3 m + 5)} = \frac{(3 m + 2) (m + 1)}{2 (3 m + 2) (3 m + 5)} = \frac{m + 1}{6 m + 10} Thus, P (m + 1) is true . B y t h e p rinciple of m athematical induction, P (n) is true for all n \in N .$

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12.5+15.8+18.11+...+1(3n-1)(3n+2)=n6n+4

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