If the sum of the first ten terms 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)

    101

  • Option 2)

    100

  • Option 3)

    99

  • Option 4)

    102

 

Answers (1)

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 ( \frac{8}{5} \right )^{2}+\left ( \frac{12}{5} \right )^{2}+\left ( \frac{16}{5} \right )^{2}+......................10\ terms\\*\\*= \frac{4^{2}}{5^{2}} \left ( 2^{2}+3^{2} +4^{2} +.........10\, \, terms \right )\\*\\*= \frac{16}{25} \left (1^2+ 2^{2}+3^{2} +4^{2} +.........11^{2}-1\, \, \right )\\*\\*=\frac{16}{25}\left ( \frac{11\times12\times23}{6}-1 \right )\\*\\*\frac{16}{25}\times 505=\frac{16}{5}\times \left ( 101 \right )


Option 1)

101

Correct

Option 2)

100

Incorrect

Option 3)

99

Incorrect

Option 4)

102

Incorrect

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