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=> 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
=> 1/2 + {1/6 + 1/12} + {1/20 + 1/30} + {1/42 + 1/56}
=> {1/2 + 1/4} + {1/12 + 1/24}
=> 3/4 + 3/24
=> 18+3/24
=> 21/24
=> 7/8

Answer is C

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Explanation:
Tn can be stated as 1/{n*(n+1)}
i.e. Tn=1/{n*(n+1)}={(n+1)-n}/{n*(n+1)}
Tn=1/n- 1/n+1
So, Sn=1- 1/n+1

here, n=7
So,

Sn= 1- 1/(7+1) = 1 - 1/8 = 7/8

IMO-C
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=>

Remember that \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{1∙2}+\frac{1}{2∙3}+\frac{1}{3∙4}+\frac{1}{4∙5}+\frac{1}{5∙6}+\frac{1}{6∙7}+\frac{1}{7∙8}\)

\(=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})\)

= \(1-\frac{1}{8}\) (because the last term of each bracket cancels out the first term of the next bracket, leaving us only the very first and very last terms)

= \(\frac{7}{8}\)

Therefore, C is the correct answer.
Answer: C
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1/2,1/6,1/12,1/20,1/30,1/42,1/56
Common term for this series is 1/n(n+1)
By partial fractions 1/n(n+1) = 1/n - 1(n+1)

Then the series becomes;
(1/1 - 1/2) , (1/2 - 1/3) , (1/3 + 1/4) ,............, (1/7 - 1/8)

Sum of the series;
(1/1 - 1/2) + (1/2 - 1/3) + (1/3 + 1/4) +............+ (1/7 - 1/8)
= 1 - 1/8
=7/8
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