If the sum of the first ten terms of the series [( 13 5 )^2 + ( 22 5 )^2 + ( 31 5 )^2 + 4^…

Question

If the sum of the first ten terms of the series $${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,$$ then m is equal to :

Accepted Answer

$${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$

$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)$$

$$ = {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)$$

$$ = {{16} \over {25}} \times 505 = {{16} \over 5} \times 101$$

$$ \Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101$$

$$ \Rightarrow m = 101.$$

Solutions · 1