13-5-15-7-17-9-12 n-12 n-3-n 32 n-3

Let P(n) be the given statement. Now, P(n)=13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)Step #160;1: #160;P(1)=13.5=115=13(2+3)Hence, #160;P(1) #160;is #1 ...

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Byju's Answer

Standard XI

Mathematics

Sequence

13.5+15.7+17....

Question

$\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}$

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Solution

Let P(n) be the given statement.
Now,
$P (n) = \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)} Step 1 : P (1) = \frac{1}{3.5} = \frac{1}{15} = \frac{1}{3 (2 + 3)} Hence, P (1) is true . Step 2 : Let P (m) be true . Then, \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 m + 1) (2 m + 3)} = \frac{m}{3 (2 m + 3)} To prove : P (m + 1) is true . T h a t i s, \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 m + 3) (2 m + 5)} = \frac{m + 1}{3 (2 m + 5)} Now, P (m) = \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 m + 1) (2 m + 3)} = \frac{m}{3 (2 m + 3)} \Rightarrow \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{(2 m + 1) (2 m + 3)} + \frac{1}{(2 m + 3) (2 m + 5)} = \frac{m}{3 (2 m + 3)} + \frac{1}{(2 m + 3) (2 m + 5)} [Adding \frac{1}{(2 m + 3) (2 m + 5)} to both side s] \Rightarrow \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{(2 m + 3) (2 m + 5)} = \frac{2 m^{2} + 5 m + 3}{3 (2 m + 3) (2 m + 5)} = \frac{(2 m + 3) (m + 1)}{3 (2 m + 3) (2 m + 5)} = \frac{m + 1}{3 (2 m + 5)} T hus, P (m + 1) is true . By the principle of mathematical induction, P (n) is true for all n \in N .$

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Similar questions

\lim_{n \to \infty} \{\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{(2 n + 1) (2 n + 3)}\}

is equal to

(a) 0
(b) 1/2
(c) 1/9
(d) 2

Q. If

n

is a positive integer, then

(\sqrt{3} + 1)^{2 n} - (\sqrt{3} - 1)^{2 n}

is:

Q. The

A . M .

of the observations

1.3.5, 3.5, 7, 9, ., (2 n - 1) (2 n + 1) (2 n + 3)

(\forall n \in N)

Q. Find the sum of n terms of the series

1.3.5 + 3.5.7 + 5.7.9 + . . . .

. The

n^{t h}

term is

(2 n - 1) (2 n + 1) (2 n + 3);

hence by the rule

S_{n} = \frac{(2 n - 1) (2 n + 1) (2 n + 3) (2 n + 5)}{4.2} + c

Q. If

z_{n} = cos \frac{π}{(2 n + 1) (2 n + 3)} + i sin \frac{π}{(2 n + 1) (2 n + 3)}

, then find the value of

lim n \to \infty (z_{1} . z_{2} . z_{3} . . . . . . . . . z_{n}) ?

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