how we can take 1/2 as a common factor (Question:327714.0)

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Prakash Kumar

· started a discussion

· 1 Months ago

how we can take 1/2 as a common factor (Question:327714.0)

I didn't understand the method you used in it. I also try to use your method but can't get the answer because I couldn't understand the method used in the question.

Question:

What is the value of the following expression?

$(\frac{1}{(2^{2} - 1)}) + (\frac{1}{(4^{2} - 1)}) + (\frac{1}{(6^{2} - 1)}) + \dots \dots + (\frac{1}{(20^{2} - 1)})$

Options:

9/19

10/19

10/21

11/21

Solution:

$(\frac{1}{(2 - 1) (2 + 1)}) + (\frac{1}{(4 - 1) (4 + 1)}) + (\frac{1}{(6 - 1) (6 + 1)}) + \dots \dots + (\frac{1}{(20 - 1) (20 + 1)})$

$(\frac{1}{1 \times 3}) + (\frac{1}{3 \times 5}) + (\frac{1}{5 \times 7}) + \dots \dots + (\frac{1}{19 \times 21})$

$\frac{1}{2} ((\frac{3 - 1}{1 \times 3}) + (\frac{5 - 3}{3 \times 5}) + (\frac{7 - 5}{5 \times 7}) + \dots \dots + (\frac{21 - 19}{19 \times 21}))$

$\frac{1}{2} (\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \dots \dots + \frac{1}{19} - \frac{1}{21})$

$\frac{1}{2} (\frac{1}{1} - \frac{1}{21}) = \frac{10}{21}$

Keerthiga

· commented

· 1 Months ago

Multiply and divide by 2, so that we could cancel some fractions

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how we can take 1/2 as a common factor (Question:327714.0)

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