Proove 13-7 - 17-11 - 111-15 - - - 14n - 14n - 3 - n 34n - 3

The correct option is A True 13· 7+17· 11+111· 15+..................+1(4n 1)(4n+3) =14(13 17+17 111+111 115+......................+1(4n 1) 1(4n+3)) ...

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Byju's Answer

Standard XII

Mathematics

Bayes' Theorem

proove 13.7 ...

Question

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

True

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False

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Solution

The correct option is A True
$\frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + . . . . . . . . . . . . . . . . . . + \frac{1}{(4 n - 1) (4 n + 3)}$
$= \frac{1}{4} (\frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \frac{1}{11} - \frac{1}{15} + . . . . . . . . . . . . . . . . . . . . . . + \frac{1}{(4 n - 1)} - \frac{1}{(4 n + 3)})$
$= \frac{1}{4} (\frac{1}{3} - \frac{1}{4 n + 3})$
$= \frac{1}{4} (\frac{4 n + 3 - 3}{3 (4 n + 3)}) = \frac{n}{3 (4 n + 3)}$
$∴ \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + . . . . . . . . . . . . . . . . . . + \frac{1}{(4 n - 1) (4 n + 3)} = \frac{n}{3 (4 n + 3)}$

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Similar questions

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

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

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proove13.7+17.11+111.15+...+1(4n−1)(4n+3)=n3(4n+3)

The correct option is A True13⋅7+17⋅11+111⋅15+..................+1(4n−1)(4n+3)=14(13−17+17−111+111−115+......................+1(4n−1)−1(4n+3))=14(13−14n+3)=14(4n+3−33(4n+3))=n3(4n+3)∴13⋅7+17⋅11+111⋅15+..................+1(4n−1)(4n+3)=n3(4n+3)

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