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Question Number 177144 by Shrinava last updated on 01/Oct/22

Calculas: ((1^2 + 2^2 + 3^2 + ... + 16^2 − 16)/(1∙3 + 2∙4 + 3∙5 + ... + 15∙17))

$$\mathrm{Calculas}: \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{2}} \:+\:...\:+\:\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{1}\centerdot\mathrm{3}\:+\:\mathrm{2}\centerdot\mathrm{4}\:+\:\mathrm{3}\centerdot\mathrm{5}\:+\:...\:+\:\mathrm{15}\centerdot\mathrm{17}} \\ $$

Answered by BaliramKumar last updated on 01/Oct/22

= ((1^2 +2^2 +3^2 +..........+16^2 − 16)/((2^2 −1)+(3^2 −1)+(4^2 −1)+...........+(16^2 −1))) = ((1^2 +2^2 +3^2 +..........+16^2 − 16)/(2^2 +3^2 +4^2 +...........+16^2 −15)) = ((1^2 +2^2 +3^2 +..........+16^2 − 16)/(1^2 +2^2 +3^2 +4^2 +...........+16^2 −15−1^2 )) = ((1^2 +2^2 +3^2 +..........+16^2 − 16)/(1^2 +2^2 +3^2 +4^2 +...........+16^2 −16)) = determinant ((1)) Answer

$$=\:\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..........+\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{4}^{\mathrm{2}} −\mathrm{1}\right)+...........+\left(\mathrm{16}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..........+\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...........+\mathrm{16}^{\mathrm{2}} −\mathrm{15}} \\ $$$$=\:\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..........+\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...........+\mathrm{16}^{\mathrm{2}} −\mathrm{15}−\mathrm{1}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +..........+\mathrm{16}^{\mathrm{2}} \:−\:\mathrm{16}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +...........+\mathrm{16}^{\mathrm{2}} −\mathrm{16}} \\ $$$$=\:\begin{array}{|c|}{\mathrm{1}}\\\hline\end{array}\:{Answer} \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa11 last updated on 02/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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