If the sum of the first ten terms of the series
 \left ( 1\frac{3}{5} \right )^{2}+\left ( 2\frac{2}{5} \right )^{2}+\left ( 3\frac{1}{5} \right )^{2}+4^{2}+\left ( 4\frac{4}{5} \right )^{2}+.........is\frac{16}{5}m

then m is equal to:

  • Option 1)

    102

  • Option 2)

    101

  • Option 3)

    100

  • Option 4)

    99

 

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}

 

 \left ( 1\frac{3}{5} \right )^{2}+\left ( 2\frac{2}{5} \right )^{2}+\left ( 3\frac{3}{5} \right )^{2}+4^{2}+\left (4\frac{4}{5} \right )^{2}----is \frac{16}{5}n

\Rightarrow \left ( \frac{8}{5} \right )^{2}+\left ( \frac{12}{5} \right )^{2}+\left ( \frac{16}{5} \right )^{2}+\left ( \frac{20}{5} \right )^{2}+ -----

\therefore\frac{1}{25} \left [ 8^{2}+12^{2}+16^{2}+20^{2}+ --- \right ]

\therefore\frac{1}{25}\times4^{}2 \left [ 2^{2}+3^{2}+4^{2}+5^{2}+--- \right ]

=\frac{16}{25} \left [ 1^{2}+2^{2}+3^{}2+-----11^{2}-1\right ]

\Rightarrow \frac{16}{25}\left [ \frac{11\times12\times23}{6}-1 \right ]=\frac{16}{5}n

\therefore 11\times46-1=5n

      506-1=5n

      505=5n

\therefore n=101

 

 


Option 1)

102

Incorrect option

Option 2)

101

Correct option

Option 3)

100

Incorrect option

Option 4)

99

Incorrect option

N neha

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