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Question 130
What is the value of the following expression?
$$\dfrac{1}{(2^2-1)}+\dfrac{1}{(4^2-1)}+\dfrac{1}{(6^2-1)}+...+\dfrac{1}{(20^2-1)}$$
Solution
$$\dfrac{1}{(2^2-1)}+\dfrac{1}{(4^2-1)}+\dfrac{1}{(6^2-1)}+...+\dfrac{1}{(20^2-1)}$$
= $$\dfrac{1}{(2+1)(2-1)}+\dfrac{1}{(4+1)(4-1)}+..+\dfrac{1}{(20+1)(20-1)}$$
= 1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/(19*21)
=1/2 * ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... +1/19 - 1/21)
=1/2 * (1 - 1/21) = 10/21
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