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What is the value of the following sum?
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
06 Sep 2019, 20:27
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Hovkial
What is the value of the following sum?
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
=> \(\frac{1}{1*2}\)+ \(\frac{1}{2*3}\)+ \(\frac{1}{3*4}\)+ \(\frac{1}{4*5}\) + \(\frac{1}{5*6}\)+ \(\frac{1}{6*7}\)+ \(\frac{1}{7*8}\)
=> (\(\frac{1}{1}-\frac{1}{2}\))+(\(\frac{1}{2}-\frac{1}{3}\))+( \(\frac{1}{3}-\frac{1}{4}\))+( \(\frac{1}{4}-\frac{1}{5}\))+(\(\frac{1}{5}-\frac{1}{6}\))+( \(\frac{1}{6}-\frac{1}{7}\))+( \(\frac{1}{7}-\frac{1}{8})=1-\frac{1}{8}\)=\(\frac{7}{8}\)
E
10 Sep 2019, 09:41
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Hovkial
What is the value of the following sum?
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\) + \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)
(A) \(\frac{1}{10}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{6}{8}\)
(D) \(\frac{5}{8}\)
(E) \(\frac{7}{8}\)
Here is general way to solve such questions if asked to find sums.
Whenever there is a "1" in the numerator, we must immediately understand that one (1) can be written as the difference of two numbers.
The above two numbers must also be two factors of the denominator.
So, we write every single one of the quantities in the sum as:
b-a/b*a (for each quantity).
Then simplify the above quantity as:
b-a/b*a = 1/a - 1/b
Adding up all the simplified quantities, we will find that all the fractions in the middle cancel each other. We will be left with the difference between the first fraction and the last fraction.
This difference will be the value of the final sum.
