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If the sum $\; \large\frac{3}{1^{2}} + \large\frac{5}{1^{2} + 2^{2}} + \large\frac{7}{1^{2} + 2^{2} +3^{2}}+----+\;$ upto 20 terms is equal to $\;\large\frac{k}{21}\;$, then k is equal to :

$(a)\;120\qquad(b)\;180\qquad(c)\;240\qquad(d)\;60$

Can you answer this question?
 
 

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General term of this series is

T_n = (2n+1)/[summation(n^2)]
 

Gives =>    T_n = 6/n(n+1)   = 6[ 1/n - 1/(n+1)]

Sum upto 20 terms =   6 [ 1- 1/21]  = 120/21  ,,,,  k=120
answered Apr 2, 2015 by Limitsofawesomeness
 
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