1-3-7-1-7-11-1-11-15- -1-4 n-14 n-3-n-34 n-3

Let P(n) : 1/3.7 + 1/7.11 + 1/11.15 + .... + 1/(4n 1)(4n + 3) = n/3(4n + 3) For n = 1 1/3.7 = 1/3(7) 1/21 = 1/21 ⇒ P(n) is true for n = 1 Let P(n) is true for ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

1/3.7+1/7.11+...

Question

$\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{n}{3 (4 n + 3)}$

Open in App

Solution

Let P(n) : $\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{n}{3 (4 n + 3)}$

For n = 1

$\frac{1}{3.7} = \frac{1}{3 (7)}$

$\frac{1}{21} = \frac{1}{21}$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{k}{3 (4 k + 3)}$ ........(1)

We have to show that,

$\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 k - 1) (4 k + 3)} + \frac{1}{(4 k + 3) (4 k + 7)} = \frac{(k + 1)}{3 (4 k + 7)}$

Now,

${\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 k - 1) (4 k + 3)}} + \frac{1}{3 (4 k + 3) (4 k + 7)}$

$= \frac{k}{3 (4 k + 3)} + \frac{1}{(4 k + 3) (4 k + 7)}$

$= \frac{1}{(4 k + 3)} [\frac{k}{3} + \frac{1}{4 k + 7}]$

$= \frac{1}{(4 k + 3)} [\frac{k (4 k + 7) + 3}{3 (4 k + 7)}]$

$= \frac{1}{(4 k + 3)} [\frac{4 k^{2} + 4 k + 3 k + 3}{3 (4 k + 7)}]$

$= \frac{1}{(4 k + 3)} [\frac{(4 k + 3) (k + 1)}{3 (4 k + 7)}]$

$\frac{1}{(4 k + 3)} [\frac{(4 k + 3) (k + 1)}{3 (4 k + 7)}]$

$= \frac{(k + 1)}{3 (4 k + 7)}$

$\Rightarrow$ p(n) is true for n = k + 1

$\Rightarrow$ p(n) is true for all n $e p s i l o n$ N by PMI

Suggest Corrections

Similar questions

Q. proove

\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{n}{3 (4 n + 3)}

Q. The value of

\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + \dots . + \frac{1}{(4 n - 1) (4 n + 3)}

{lim}_{n \to \infty} (\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . n terms) =

\frac{1}{1.5} + \frac{1}{5.9} + \dots . + \frac{1}{(4 n - 3) (4 n + 1)} =

Q. The sum of the

n

terms of the series

3 + 8 + 22 + 72 + 266 + 1036 + \dots

Join BYJU'S Learning Program

13.7+17.11+111.15+....+1(4n−1)(4n+3)=n3(4n+3)

$\frac{1}{3.7} + \frac{1}{7.11} + \frac{1}{11.15} + . . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{n}{3 (4 n + 3)}$