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)

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

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