Question

please answer this question .....
please.....

Answer 1

Hi
${1}^{2} + {2}^{2} + {3}^{2} + ... + {19}^{2} = \frac{19 \times 20 \times (19 \times 2 + 1)}{ 6}$
Now
${1}^{2} + {3}^{2} + {5}^{2} + ... + {19}^{2} = above \: value - ( {2}^{2} + {4}^{2} + {6}^{2} + ... + {18}^{2} )$

But
${2}^{2} + {4}^{2} + {6}^{2} + ... + {18}^{2} = \\ {( 2)}^{2} + {(2 \times 2)}^{2} + {(2 \times 3)}^{2} + ... + {(2 \times 9)}^{2} = \\ {2}^{2} ( {1}^{2} + {2}^{2} + {3}^{2} + .. + {9}^{2} ) = \\ {2}^{2} \times \frac{9 \times 10 \times (2 \times 9 + 1)}{6}$
I believe you got my point and you can finish it.
Hope this helps.

Answer 2

Solution:-

given :-
${1 }^{2} + {2}^{2} + { 3}^{2} ..... {x}^{2} = \frac{x(x + 1)(2x + 1)}{6} \\ {1}^{2} + {2}^{2} + {3}^{2} ... {19}^{2} - ({2}^{2} + {4}^{2} + {6}^{2} ....... {18}^{2} ) = {1}^{2} + {3}^{2} + {5}^{2} .... {19}^{2} \\ \frac{19 \times 20 \times 39}{6} - {2}^{2} ( \frac{9 \times 10 \times 19}{6}) \\ 2470 - 4 \times 285 \\ 2470 - 1140 \\ 1330ans$