A) 120
B) 180
C) 240
D) 60
Correct Answer: A
Solution :
nth term of given series is\[\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}\] \[=\frac{6}{n(n+1)}\] Let \[{{n}^{th}}\]term,\[{{a}_{n}}=6\left[ \frac{1}{n}-\frac{1}{n+1} \right]\] Sum of 20 terms, \[{{S}_{20}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\]?\[+{{a}_{20}}\] \[{{S}_{20}}=3\left( \frac{1}{1}-\frac{1}{2} \right)+6\left( \frac{1}{2}-\frac{1}{3} \right)+6\left( \frac{1}{3}-\frac{1}{4} \right)+...\] \[+6\left( \frac{1}{18}-\frac{1}{19} \right)+6\left( \frac{1}{19}-\frac{1}{20} \right)+6\left( \frac{1}{20}-\frac{1}{21} \right)\] \[{{S}_{20}}=\left[ \left( 1-\cancel{\frac{1}{2}} \right)+\left( \cancel{\frac{1}{2}}-\cancel{\frac{1}{3}} \right)+\left( \cancel{\frac{1}{3}}-\cancel{\frac{1}{4}} \right)+... \right.\] \[+\left. \left( \cancel{\frac{1}{18}-}\cancel{\frac{1}{19}} \right)+\left( \cancel{\frac{1}{19}}-\cancel{\frac{1}{20}} \right)+\left( \cancel{\frac{1}{20}}-\frac{1}{21} \right) \right]\] \[{{S}_{20}}=6\left( 1-\frac{1}{21} \right)=\frac{120}{21}\] ?(1) Given that \[{{S}_{20}}=\frac{k}{21}\] ?(2) On comparing (1) and (2), we get k = 120
You need to login to perform this action.
You will be redirected in
3 sec
