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  • Please,please help me - Sequence and series - JEE Main-2

If the sum  \frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+.....+   up to 20 terms is equal to \frac{k}{21}, then k is equal to :

  • Option 1)

    120

  • Option 2)

    180

  • Option 3)

    240

  • Option 4)

    60

 

Answers (2)

best_answer

As we learnt in 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )
 

- wherein

Sum of  squares of first n natural numbers

1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}

\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+.....+   20 terms

T_{n}=\frac{(2n+1)\times 6}{n(n+1)(2n+1)}= \frac{6}{n(n+1)}

\sum T_{n}=S_{n}=\sum 6\left [ \frac{1}{n(n+1)} \right ]

           =\sum 6\left [ \frac{1}{n}-\frac{1}{n+1} \right ]

\Rightarrow\ S_{n}= 6\left ( 1-\frac{1}{n+1} \right )

                  =6\left [ 1-\frac{1}{n+1} \right ]= \frac{6n}{n+1}

Put n=20

    \frac{6\times 20}{21}=\frac{k}{21} \: (given)

k=120


Option 1)

120

This option is correct

Option 2)

180

This option is incorrect

Option 3)

240

This option is incorrect

Option 4)

60

This option is incorrect

Posted by

prateek

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